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A problem with logarithms If $\log(a+b+c)=\log(a) + \log(b) + \log(c)$, prove that $$\log\left(\frac{2a}{1-a^2} +\frac{2b}{1-b^2} +\frac{2c}{1-c^2}\right) = \log\left(\frac{2a}{1-a^2}\right)+ \log\left(\frac{2b}{1-b^2}\right)+\log\left(\frac{2c}{1-c^2}\right) $$
|
Assuming $a,b,c>0$, Let:
$a =\tan\alpha $
$b =\tan\beta $
$c =\tan\gamma$
Inserting it in the given relation of the question, we get:-
$$\log(a+b+c)=\log(a) + \log(b) + \log(c)$$
$$=> a+b+c= abc => \tan\alpha +\tan\beta+\tan\gamma=\tan\alpha\tan\beta\tan\gamma$$
$$=> \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=-\tan\gamma=\tan(\pi-\gamma) $$
Taking $ \alpha ,\beta,\gamma $ to be in the first quadrant, we get $\alpha+\beta+\gamma=\pi => 2\alpha+2\beta+2\gamma=2\pi $
Notice that:
$\tan2\alpha+\tan2\beta+\tan2\gamma-\tan2\alpha\tan2\beta\tan2\gamma =0$
On transposition, taking log, and then subsequently substituting back our inital substitutions, we get: $$\log\left(\frac{2a}{1-a^2} +\frac{2b}{1-b^2} +\frac{2c}{1-c^2}\right) = \log\left(\frac{2a}{1-a^2}\right)+ \log\left(\frac{2b}{1-b^2}\right)+\log\left(\frac{2c}{1-c^2}\right)$$
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|
How to simplify Boolean Expression $\bar B + \bar C (B + A)$ I trying to figure out how $ \bar B + \bar C (B + A)$ simplifies to
$ \bar B + \bar C$.
|
It's usually easier to figure these things out with Karnaugh maps. After playing around, we see that:
\begin{align*}
\overline B + \overline C(B + A)
&= \overline B(1) + \overline C(B + A) \\
&= \overline B(1 + \overline C) + \overline C(B + A) \\
&= \overline B + \overline B ~ \overline C + \overline C(B + A) \\
&= \overline B + \overline C(\overline B + B + A) \\
&= \overline B + \overline C(1 + A) \\
&= \overline B + \overline C(1) \\
&= \overline B + \overline C
\end{align*}
as desired.
|
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Cyclic Equation. Prove that: $\small\frac { a^2(b-c)^3 + b^2(c-a)^3 + c^2(a-b)^3 }{ (a-b)(b-c)(c-a) } = ab + bc + ca$? This is how far I got without using polynomial division:
\begin{align}
\tiny
\frac { a^{ 2 }(b-c)^{ 3 }+b^{ 2 }(c-a)^{ 3 }+c^{ 2 }(a-b)^{ 3 } }{ (a-b)(b-c)(c-a) }
&\tiny=\frac { { a }^{ 2 }\{ { b }^{ 3 }-{ c }^{ 3 }-3bc(b-c)\} +{ b }^{ 2 }\{ { c }^{ 3 }-{ a }^{ 3 }-3ca(c-a)\} +c^{ 2 }\{ a^{ 3 }-b^{ 3 }-3ab(a-b)\} }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { a }^{ 2 }({ b }^{ 3 }-{ c }^{ 3 })-3{ a }^{ 2 }bc(b-c)+{ b }^{ 2 }({ c }^{ 3 }-{ a }^{ 3 })-3ab^{ 2 }c(c-a)+c^{ 2 }(a^{ 3 }-b^{ 3 })-3abc^{ 2 }(a-b) }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { a }^{ 2 }({ b }^{ 3 }-{ c }^{ 3 })+{ b }^{ 2 }({ c }^{ 3 }-{ a }^{ 3 })+c^{ 2 }(a^{ 3 }-b^{ 3 })-3{ a }^{ 2 }bc(b-c)-3ab^{ 2 }c(c-a)-3abc^{ 2 }(a-b) }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { a }^{ 2 }({ b }^{ 3 }-{ c }^{ 3 })+{ b }^{ 2 }({ c }^{ 3 }-{ a }^{ 3 })+c^{ 2 }(a^{ 3 }-b^{ 3 })-3{ a }bc\{ (b-c)+(c-a)+(a-b)\} }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { a }^{ 2 }({ b }^{ 3 }-{ c }^{ 3 })+{ b }^{ 2 }({ c }^{ 3 }-{ a }^{ 3 })+c^{ 2 }(a^{ 3 }-b^{ 3 }) }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { -a }^{ 3 }(b^{ 2 }-c^{ 2 })+{ a }^{ 2 }({ b }^{ 3 }-{ c }^{ 3 })-b^{ 2 }c^{ 2 }(b-c) }{ (a-b)(b-c)(c-a) } \\
&\tiny=\frac { { -a }^{ 3 }(b+c)+{ a }^{ 2 }({ b }^{ 2 }+{ bc+c }^{ 2 })-b^{ 2 }c^{ 2 } }{ (a-b)(c-a) }
\end{align}
Would it be possible to solve this answer easily without direct polynomial division? By the use of some known identities, perhaps?
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If you don't feel like being clever, brute forcing will also work: the numerator
\begin{align*}
&=a^2(b-c)^3 + b^2(c-a)^3 + c^2(a-b)^3\\
&=a^2[(b-a)+(a-c)]^3+b^2(c-a)^3+c^2(a-b)^3\\
&=a^2(b-a)^3+3a^2(b-a)(a-c)(b-c)+a^2(a-c)^3+b^2(c-a)^3+c^2(a-b)^3\\
&=-(c-a)^3(a-b)(a+b)+(a-b)^3(c-a)(c+a)+3a^2(b-a)(a-c)(b-c)
\end{align*}
which is $(a-b)(b-c)(c-a)$ multiplied with
\begin{align*}
&\quad-\frac{(c-a)^2(a+b)}{b-c}+\frac{(a-b)^2(c+a)}{b-c}+3a^2\\
&=\left[-\frac{(c-a)^2a}{b-c}+\frac{(a-b)^2a}{b-c}\right]+\left[-\frac{(c-a)^2b}{b-c}+\frac{(a-b)^2c}{b-c}\right]+3a^2\\
&=a(-2a+b+c)+(-a^2+bc)+3a^2=ab+ac+bc.
\end{align*}
The main thing is that you break the problem into chewable sizes.
|
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|
Find all pairs of positive integers $(m,n)$ such that $2^{m+1}+3^{n+1}$ is a perfect square Find all pairs of positive integers $(m,n)$ such that $2^{m+1}+3^{n+1}$ is a perfect square
My attempt so far
Any perfect square is $0,1$ in mod 4, so $n+1$ must be even :
$$2^{m+1}+3^{2r}=k^2$$
Rearranging and factoring
$$2^{m+1}=(k+3^r)(k-3^r)$$
That gives two equations
$$2^a=k+3^r\\2^b=k-3^r\\a+b=m+1;a>b$$
However I am not able to conclude, subtracting/adding... these equations is not giving me anything useful. Any help is appreciated... Thanks!
|
All perfect squares are either $0$ or $1 \pmod 3$ so $m+1$ is also even. We now have:
$$2^{2s} + 3^{2r} = k^2$$
Or, equivalently:
$$(2^s)^2 + (3^r)^2 = k^2$$
Wich is a pythagorean triple.
Since $(2^s,3^r) = 1$ then there are integers $p,q$ so that:
$$\begin{cases}2pq = 2^s & \\
p^2 - q^2 = 3^r & \\
p^2 + q^2 = k &
\end{cases}$$
The first equation means that $pq = 2^{s-1}$ thus $p = 2^a$ and $q = 2^b$ for some integers $a,b$ such that $a + b = s-1$
But then because of the second equation then if both $a$ and $b$ are nonzero $4 \mid p^2 - q^2 = 3^r$ wich is false.
If both $a$ and $b$ are zero then $3^r = 0$ wich is also impossible.
Therefore $a > 0$ and $b = 0$
We must now fint the values of $a$ for wich $2^{2a} - 1 = 3^r$
Or equivalently, $4^a - 1 = 3^r$. Checking modulo $4$ we deduce that $r$ is odd.
For $r = 1$ we have $a = 1$.
For $r > 2$ by checking modulo $9$ we deduce that $3\mid a$, thus $4^{3a'} - 1 = 3^{2r'+1}$
So $4^{3a'} = 3^{2r'+1} + 1$ with $3\nmid 2r'+1$
Or equivalently, $64^{a'} = 3\cdot 9^{r'} + 1$.
By checking modulo $64$ then $LHS = 0$ but $RHS\in \{4,28,52,12,36,60,20,44\}$.
Therefore, the only solution is indeed $(m,n,k) = (1,1,5)$.
|
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|
Did I do this Continuous Probability Problem Correctly? I'm new to evaluating continuous probability density functions. I'd like someone to check my work, please.
Problem:
Suppose $X$ has density $f(x) = c/x^6$ for $x>1$ and $f(x) = 0$
otherwise, where $c$ is a constant.
*
*Find $c$.
*Find $E(X)$.
*Find $Var(X)$.
Solution:
*
*The big rule we'll refer to is $\int_{-\infty}^\infty f(x)dx = 1$. Since $f(x) = 0$ for $x \leq 1$, it is clear that we only have to consider
$$\int_1^\infty f(x) dx = \int_1^\infty \frac{c}{x^6} dx = c \cdot \int_1^\infty \frac{1}{x^6} dx = c \cdot \left[ \frac{-1}{5x^5} \right]^\infty_1 = c \cdot \lim_{x \to \infty} \left[ \frac{1}{5} - \frac{1}{5x^5} \right] = \frac c 5$$
And since it must be that $\int_{-\infty}^\infty f(x)dx = 1$, we must have that $\frac c 5 = 1$ so $c=5$.
*We may calculate $E(X) = \int_{-\infty}^\infty xf(x)dx$. Again, since $f(x) = 0$ for $x \leq 1$, we want to calculate
$$\int_1^\infty xf(x) dx = \int_1^\infty x\frac{5}{x^6}dx = 5\int_1^\infty \frac{1}{x^5}dx = 5 \left[ \frac{-1}{4x^4}\right]^\infty_1 = 5 \lim_{x\to\infty} \left[ \frac 1 4 - \frac{1}{4x^4}\right] = \frac 5 4$$
*We calculate $Var(X) = E(X^2) - (E(X))^2$. We know $(E(X))^2 = \left(\frac{5}{4}\right)^2 = \frac {25}{16}$. We need $E(X^2)$, which we may calculate with:
$$\int_1^\infty x^2f(x) dx = 5\int_1^\infty \frac{1}{x^4}dx = 5 \left[ \frac{-1}{3x^3} \right]_1^\infty = 5 \lim_{x \to \infty} \left[ \frac 1 3 - \frac{1}{3x^3} \right] = \frac 5 3$$
So $Var(X) = \frac 5 3 - \frac{25}{16} = \frac{25}{15} - \frac{25}{16} = 25 \left( \frac{1}{15} - \frac{1}{16}\right) = 25 \left( \frac{16}{15\cdot 16} - \frac{15}{16 \cdot 15}\right) = \frac{25}{16 \cdot 15} = \frac 5 {48}$.
|
Your reasoning and execution are perfectly correct.
|
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|
Simple Lagrange Multiplyers Problem Can anyone please help me with the following:
Find the stationary values of $u=x^2+y^2$ subject to the constraint $t(x,y) = 4x^2 + 5xy + 3y^2 = 9$.
The answer is given as $u = 9$ and $x = \pm 3/\sqrt2$ and $y = \mp 3/\sqrt2$
I do the following:
Let $g(x,y) = 4x^2 + 5xy + 3y^2 - 9 = 0$
$$\begin{align*}
\partial u/\partial x &= 2x \\
\partial u/\partial y &= 2y \\
\partial g/\partial x &= 8x + 5y \\
\partial g/\partial y &= 5x + 6y
\end{align*}$$
So solve the system:
$$\begin{align*}
2x + k(8x + 5y) &= 0\\
2y + k(5x + 6y) &= 0\\
4x^2 + 5xy + 3y^2 - 9 &= 0
\end{align*}$$
And this system does not solve to give the stated answer.
Where have I erred, or is the book answer (or question?) wrong?
Thanks,
Mitch.
|
I haven't worked out a numerical answer by hand:
if I eliminate k from:
$$\begin{align*}
2x + k(8x + 5y) &= 0\\
2y + k(5x + 6y) &= 0\\
\end{align*}$$
I get:
$$\begin{align*}
2y - \frac{2x(5x+6y)}{8x+5y} &= 0\\
\end{align*}$$
And if one substitutes in
$$\begin{align*}
x=\frac{3}{\sqrt{2}}\\
y=\frac{-3}{\sqrt{2}}
\end{align*}$$
One would expect to get zero (if the supplied answers are correct) but one gets
$$\begin{align*}
-2\sqrt{2}
\end{align*}$$
Mitch.
|
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|
Prove that the integral of $\sin^2(x)/(5+3\cos(x))$ from $0$ to $2\pi$ is $2\pi/9$ I'm not really unsure of how to approach this problem. I was thinking of reparametrizing the sin and the cos to its exponential form but I realize that it becomes even messier and leads sort of nowhere. There are no singularities for this function f(x) I believe, so there's not really a way to use the Residue theorem either. Can anyone give me some help on this?
|
Suppose we seek to evaluate
$$\int_0^{2\pi} \frac{\sin^2 x}{5+3\cos x} dx.$$
Put $z = \exp(ix)$ so that $dz = i\exp(ix) \; dx$ and hence
$\frac{dz}{iz} = dx$ to obtain
$$\int_{|z|=1}
\frac{(z-1/z)^2/4/(-1)}{5+3/2(z+1/z)} \frac{dz}{iz}
\\ = -\int_{|z|=1}
\frac{(z-1/z)^2}{20+6(z+1/z)} \frac{dz}{iz}
\\ = -\int_{|z|=1}
\frac{z^2-2+1/z^2}{20+6z+6/z} \frac{dz}{iz}
\\ = -\int_{|z|=1}
\frac{z^4-2z^2+1}{20z^2+6z^3+6z} \frac{dz}{iz}
\\ = -\frac{1}{i} \int_{|z|=1}
\frac{1}{z^2} \frac{z^4-2z^2+1}{6z^2+20z+6} \; dz.$$
There are three poles, one at $z=-3$, another one at $z=-1/3$ and
another one at $z=0.$ Only the latter two contribute.
The pole at $z=-1/3$ is simple and hence the residue is
$$\left.\frac{1}{z^2}
\frac{z^4-2z^2+1}{12z+20}\right|_{z=-1/3}
= \frac{4}{9}.$$
The remaining contribution is from
$$\int_{|z|=1}
\frac{1}{z^2} \frac{1}{6z^2+20z+6} \; dz
=\frac{1}{6}
\int_{|z|=1}
\frac{1}{z^2} \frac{1}{z^2+10/3z+1} \; dz
\\ =\frac{1}{6}
\int_{|z|=1}
\frac{1}{z^2} \sum_{q\ge 0} (-1)^q z^q (z+10/3)^q \; dz.$$
The only contribution in the series is from $q=1$ and it is
$$-\frac{1}{6} 10/3 = -\frac{5}{9}.$$
Collecting everything we get
$$-\frac{1}{i} \times 2\pi i\times
\left(\frac{4}{9}-\frac{5}{9}\right)
= \frac{2\pi}{9}.$$
|
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How does $-\frac{1}{x-2} + \frac{1}{x-3}$ become $\frac{1}{2-x} - \frac{1}{3-x}$ I'm following a solution that is using a partial fraction decomposition, and I get stuck at the point where $-\frac{1}{x-2} + \frac{1}{x-3}$ becomes $\frac{1}{2-x} - \frac{1}{3-x}$
The equations are obviously equal, but some algebraic manipulation is done between the first step and the second step, and I can't figure out what this manipulation could be.
The full breakdown comes from this solution
$$
\small\begin{align}
\frac1{x^2-5x+6}
&=\frac1{(x-2)(x-3)}
=\frac1{-3-(-2)}\left(\frac1{x-2}-\frac1{x-3}\right)
=\bbox[4px,border:4px solid #F00000]{-\frac1{x-2}+\frac1{x-3}}\\
&=\bbox[4px,border:4px solid #F00000]{\frac1{2-x}-\frac1{3-x}}
=\sum_{n=0}^\infty\frac1{2^{n+1}}x^n-\sum_{n=0}^\infty\frac1{3^{n+1}}x^n
=\bbox[4px,border:1px solid #000000]{\sum_{n=0}^\infty\left(\frac1{2^{n+1}}-\frac1{3^{n+1}}\right)x^n}
\end{align}
$$
Original image
|
Each of the terms was multiplied by $\frac{-1}{-1}$, which is really equal to $1$, so it's a "legal" thing to do:
$-\dfrac{1}{x - 2} + \dfrac{1}{x - 3}$
$ = -\dfrac{(-1)1}{(-1)(x - 2)} + \dfrac{(-1)1}{(-1)(x - 3)}$
$ = -\dfrac{-1}{2 - x} + \dfrac{-1}{3 - x}$
$ = \dfrac{1}{2 - x} - \dfrac{1}{3 - x} $
|
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How to compute $\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}}$? I have to compute the series $\displaystyle\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}}$.
$$\displaystyle\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}} = \sum_{n=0}^{\infty}{\frac{3^n\frac{1}{2}}{n!}} + \sum_{n=0}^{\infty}{\frac{3^nn}{n!}} = \frac{e^3}{2} + \sum_{n=0}^{\infty}{\frac{3^nn}{n!}},$$ but how to compute the $\displaystyle\sum_{n=0}^{\infty}{\frac{3^nn}{n!}}$?
|
$$\begin{align}
\sum_{n=0}^\infty\frac{3^nn}{n!} & =0+\sum_{n=1}^\infty\frac{3^n}{(n-1)!}\\
& =\sum_{n=0}^\infty\frac{3^{n+1}}{n!}\\
& =3e^3.\\
\end{align}$$
|
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Not understanding solution to $\large \int_{|z-2|=2} \frac {5z+7}{z^2+2z-3}dz$ computation Not understanding solution to $\large \int_{|z-2|=2} \frac {5z+7}{z^2+2z-3}dz$ computation.
What was shown in class: $\large \int_{|z-2|=2} \frac {5z+7}{z^2+2z-3}dz=\large \int_{|z-2|=2} \frac {5z+7}{(z+3)(z-1)}=\cdots= \int_{|z-2|=2} \frac {2}{z+3}dz+\int_{|z-2|=2} \frac {3}{z-1}dz $
And then it is said that $\large \int_{|z-2|=2} \frac {2}{z+3}dz=0$ while $\large \int_{|z-2|=2} \frac {3}{z-1}dz=3 \cdot 2 \cdot \pi \cdot i$.
Can you please help me understand the last step? What's the difference bewteen the integrals?
|
Let $D=\{z: |z-2|\leq2\}$, thus $\partial D=\{|z-2|=2\}$. Since $f(z)=2/(z+3)$ is clearly analytic in $D$, then by Cauchy's Theorem $$ \int_{\partial D} f(z)dz=0.$$
However, if $g(z)=3/(z-1)$, since $1 \in D$, by Cauchy's Integral Formula, letting $g(z)=h(z)/(z-1)$ with $h\equiv 3$ we get
$$
\int_{\partial D} g(z)dz=\int_{\partial D} \frac{h(z)}{z-1} dz= 2\pi i \cdot h(1) = 2\pi i \cdot 3 = 6\pi i
$$
|
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|
Limit of functions involving trigonometry as n approaches infinity
By graphing these functions, I know that P(n) approaches pi as n tends towards infinity. However, is there a mathematical way for proving this?
I am doing a maths exploration on Archimedes' approximation of pi and those are the formulas that I derived for polygons that fit perfectly outside (the first function) and inside (the second function) of a circle with a diameter of 1 (and they work!).
I have also taken the derivative of the functions to show that they are monotone increasing (for the second function) and decreasing (for the first function).
|
I think that Taylor series could be a simple solution.
Considering that, for small $x$, $$\tan(x)=x+\frac{x^3}{3}+O\left(x^4\right)$$ $$n \tan\big(\frac{\pi}n\big)=n \Big(\frac{\pi }{n}+\frac{\pi ^3}{3 n^3}+\cdots)=\pi+\frac{\pi ^3}{3 n^2}+\cdots$$ For the second (without using the good hint user222031 provided in comments) $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$\cos\big(\frac{2\pi}n\big)=1-\frac{2 \pi ^2}{n^2}+\frac{2 \pi ^4}{3
n^4}+\cdots$$ $$\frac 12 -\frac 12 \cos\big(\frac{2\pi}n\big)=\frac{\pi ^2}{n^2}-\frac{\pi ^4}{3 n^4}+\cdots=\frac{\pi ^2}{n^2}\big(1-\frac{\pi ^2}{3n^2}+\cdots\big)$$ $$\sqrt{\frac 12 -\frac 12 \cos\big(\frac{2\pi}n\big)}=\frac{\pi }{n}\sqrt{1-\frac{\pi ^2}{3n^2}+\cdots}=\frac{\pi }{n}\big(1-\frac{\pi ^2}{6n^2}+\cdots\big)$$ $$n\sqrt{\frac 12 -\frac 12 \cos\big(\frac{2\pi}n\big)}=\pi\big(1-\frac{\pi ^2}{6n^2}+\cdots\big)$$
For sure, this be can be done faster using user222031's hint since $$\sqrt{\frac 12 -\frac 12 \cos\big(\frac{2\pi}n\big)}=\sin\big(\frac{\pi}n\big)$$ and since $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ then $$n\sin\big(\frac{\pi}n\big)=n\Big(\frac{\pi }{n}-\frac{\pi ^3}{6 n^3}+\cdots\Big)=\pi\big(1-\frac{\pi ^2}{6n^2}+\cdots\big)$$
|
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|
Minimum value of trigonometric equation Find the minimum value of the expression $$y=\frac{16-8\sin^{2} 2x +8\cos^{4} x}{\sin^{2} 2x} .$$ When I convert the expression completely into $2x$, cross multiply and make the discriminant of the quadratic equation greater than $0$, I get the minimum value $-\infty$. I know it is wrong, but why?
|
$$y=\frac{16-8\sin^{2} 2x +8\cos^{4} x}{\sin^{2} 2x}$$
or $$y=\frac{16}{4}*\sec^2 x \csc^2 x-8 +\frac{8}{4}\frac{\cos^{4} x}{\sin^2 x* \cos^2 x}$$
or
$$y=4(\tan^2 x +1)(\cot^2 x +1)-8+2 \cot^2 x$$
$$y=4(\tan^2 x + \cot^2 x + 2)-8+2 \cot^2 x$$
Which after simplifying gives,
$$y=4\tan^2 x + 6\cot^2 x$$
Edit: as user suggested the answer in comment is not valid. But one here can easily use AM-GM inequality to reach at correct answer
|
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"url": "https://math.stackexchange.com/questions/1280639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Computing a limit similar to the exponential function I want to show the following limit:
$$
\lim_{n \to \infty}
n
\left[
\left( 1 - \frac{1}{n} \right)^{2n}
- \left( 1 - \frac{2}{n} \right)^{n}
\right]
= \frac{1}{e^{2}}.
$$
I got the answer using WolframAlpha, and it seems to be correct numerically, but I am having trouble proving the result. My first instinct was to write the limit as
$$
\lim_{n \to \infty}
\frac
{
\left( 1 - \frac{1}{n} \right)^{2n}
- \left( 1 - \frac{2}{n} \right)^{n}
}
{1/n}.
$$
Then, I tried applying l'Hopital's rule, and I got
$$
\lim_{n \to \infty}
\frac
{
\left( 1 - \frac{1}{n} \right)^{2n}
\left( 2 \log\left( 1 - \frac{1}{n} \right) + \frac{2}{n-1} \right)
-
\left( 1 - \frac{2}{n} \right)^{n}
\left( \log\left( 1 - \frac{2}{n} \right) + \frac{2}{n-2} \right)
}
{-1/n^{2}}.
$$
This does not seem to have gotten me anywhere. My second attempt was to use the binomial theorem:
$$
\begin{align*}
n
\left[
\left( 1 - \frac{1}{n} \right)^{2n}
- \left( 1 - \frac{2}{n} \right)^{n}
\right]
&
=
n
\left[
\sum_{k=0}^{2n} \binom{2n}{k} \frac{(-1)^{k}}{n^{k}}
- \sum_{k=0}^{n} \binom{n}{k} \frac{(-1)^{k} 2^{k}}{n^{k}}
\right]
\\ &
=
\sum_{k=2}^{n} \left[ \binom{2n}{k} - \binom{n}{k} 2^{k} \right] \frac{(-1)^{k}}{n^{k-1}}
+ \sum_{k=n+1}^{2n} \binom{2n}{k} \frac{(-1)^{k}}{n^{k-1}}.
\end{align*}
$$
At this point I got stuck again.
|
I would do the transform
$$\left(1-\frac{1}{n}\right)^{2n}=e^{2n\log\left(1-\frac{1}{n}\right)}$$
then use the second order Taylor expansion
$$\log(1+x)\approx x-\frac{x^2}{2}$$
and similarly for the other term, obtaining
$$
n\left(e^{-2-\frac{1}{n}}-e^{-2-\frac{2}{n}}\right)=e^{-2-\frac{2}{n}}\cdot\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1282610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
}
|
find equality between linear spans $$U = Sp\{(2,5,-4,-10), (1,1,1,1),(1,0,3,5), (0,2,-4,-8)\}$$
$$ W = Sp\{(1,-2,7,13), (3,1,7,11), (2,1,4,6) \}$$
two questions:
*
*prove that $U = W$
*find the values of the $a \in \mathbb{R}$ where the vector $v=(a,a-6,4a-3,6a-1)$ belongs to $U$
i thought about solving it with a matrix but i wasn't sure how to approach it maybe something like
$$ \begin{pmatrix}
2 & 1 & 1 & 0 &| & 1 & 3 & 2 \\
5 & 1 & 0 & 2 &| & -2 &1 &1\\
-4 & 1 & 3 & -4&|&7& 7& 4 \\
-10 & 1 & 5 & -8&| & 13 &11 &6
\end{pmatrix} $$
and to use row reduction some how to reach zero maybe?
also what about the second question was lost there also
|
Let
\begin{align*}
A &=
\begin{bmatrix}
2&1&1&0\\
5&1&0&2\\
-4&1&3&-4\\
-10&1&5&-8
\end{bmatrix}
&
B&=
\begin{bmatrix}
1&3&2\\
-2&1&1\\
7&7&4\\
13&11&6
\end{bmatrix}
\end{align*}
Note the equation $AX=B$ has a (non-unique!) solution
$$
X=
\begin{bmatrix}
-1&-2/3&-1/3\\
3&13/3&8/3\\
0&0&0\\
0&0&0
\end{bmatrix}
$$
Also note the equation $BY=A$ has a (also non-unique!) solution
$$
Y=
\begin{bmatrix}
-13/7&-2/7&1/7&-6/7\\
9/7&3/7&2/7&2/7\\
0&0&0&0
\end{bmatrix}
$$
Do you see how these equations relate to your problem?
That $AX=B$ has a solution implies that $U\subseteq W$ and that $BY=A$ has a solution implies that $W\subseteq U$. Hence $U=W$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1284045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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|
Simplifying quartic complex function in terms of $\cos nx$ $$z= \cos(x)+i\sin(x)\\
3z^4 -z^3+2z^2-z+3$$
How would you simplify this in terms of $\cos(nx)$?
|
$$z= \cos(x)+i\sin(x)=e^{xi}$$
$$3z^4 -z^3+2z^2-z+3$$
So we got:
$$3(e^{xi})^4-(e^{xi})^3+2(e^{xi})^2-(e^{xi})+3=$$
$$3e^{4xi}-e^{3xi}+2e^{2xi}-e^{xi}+3=$$
$$(e^{ix}+e^{2ix}+1)(-4e^{ix}+3e^{2ix}+3)=$$
$$2e^{2ix}(2\cos(x)+1)(3\cos(x)-2)=$$
$$2e^{2ix}\left(6\cos^2(x)-\cos(x)-2\right)=$$
$$(2(\cos(2x)+\sin(2x)i))\left(6\cos^2(x)-\cos(x)-2\right)=$$
$$(2(\cos(2x)+((2(\cos\left(\frac{\pi}{2}-x\right)))\cos(x))i))\left(6\cos^2(x)-\cos(x)-2\right)=$$
$$\left(2\cos(2x)+\left(\left(2\cos\left(\frac{\pi}{2}-x\right)\right)\cos(x)\right)i\right)\left(6\cos^2(x)-\cos(x)-2\right)=$$
$$\left(2\cos(2x)+\left(2\cos(x)\cos\left(\frac{\pi}{2}-x\right)\right)i\right)\left(6\cos^2(x)-\cos(x)-2\right)=$$
$$\left(2\cos(2x)+2\cos(x)\cos\left(\frac{\pi}{2}-x\right)i\right)\left(6\cos^2(x)-\cos(x)-2\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1286191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
|
Probability distribution of number of columns that has two even numbers in a chart
We distribute numbers $\{1,2,...,10\}$ in random to the following chart:
Let $X$ be the number of columns that has two even numbers.
What is the distribution of $X$?
My attempt:
$|\Omega|=10!$
$P(x=0)=\frac {(5!)^2\cdot 2 ^5}{10!}=\frac 8 {63}$ Explanation: Arrange the even and odd numbers, choose which one is up 5 times.
I'm having trouble with the rest, for $P(x=1)$: choose two even numbers and a column $5\binom 5 2$, this force a column with odd numbers so choose them and a column $4\binom 5 2$, now we're left with arranging the rest like before: $(3!)^2\cdot 2^3$ so:
$P(x=1)=\frac{5\binom 5 24\binom 5 2 (3!)^2\cdot 2^3}{10!}= \frac {10}{63}$ which is too small to be right, and if I use the same reasoning $P(x=2)$ is too small again so they don't add up to 1.
|
Let us grind it out.
There are five odd and five even. So we might as well assume that we use five $0$'s and five $1$'s. Let $X$ be the number of columns with two $0$'s. We want the probability distribution of $X$.
The random variable $X$ can only take on the values $0$, $1$, and $2$. So we really have only two probabilities to compute, since if we know two of $\Pr(X=0)$, $\Pr(X=1)$, and $\Pr(X=2)$ then the third is easy.
What is the probability that $X=2$? For that to happen, we need $2$ or $3$ $0$'s in the top row. Let us deal with $2$ $0$'s and then, because of symmetry, multiply by $2$.
There are $\binom{10}{5}$ equally likely ways to place the $0$'s. Of these ways, $\binom{5}{2}\binom{5}{3}$ have $2$ $0$'s in the top row. For the locations in the top row can be chosen in $\binom{5}{2}$ ways, and for each of these ways the locations in the bottom row can be chosen in $\binom{5}{3}$ ways. So the probability of $2$ $0$'s in the top row is $\frac{\binom{5}{2}\binom{5}{3}}{\binom{10}{5}}$.
Given there are $2$ $0$'s in the top row, what is the probability of $2$ columns of $0$'s? Without loss of generality we may assume the $0$'s in the top row occupy the first $2$ positions. We want the probability that $2$ of the bottom $0$'s are in the first $2$ positions. Of the $\binom{5}{3}$ ways that the bottom $0$'s can be arranged, $3$ have the desired property. So
$$\Pr(X=2)=2\cdot\frac{\binom{5}{2}\binom{5}{3}}{\binom{10}{5}}\cdot \frac{3}{\binom{5}{3}}.$$
It's your turn! I suggest going after $\Pr(X=0)$. This can happen in two types of ways: all the $0$'s are in one of the rows or it is a 4-1 (or 1-4) distribution and there is no match.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1286415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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|
How to prove that $2^{n+2}+3^{2n+1}$ is divisible by 7 using induction? I want to prove that $2^{n+2}+3^{2n+1}$ is divisible by 7 using induction. My first step is replace $n$ with $1$.
*
*$2^{1+2}+3^{2(1)+1}$
*$2^3+3^3$
*$8+27$
*$35 = 7\times 5$
The next step is assume that $2^{n+2}+3^{2n+1}$ is divisible by 7. And the last step is to prove that $2^{(n+1)+2}+3^{2(n+1)+1}$.
*
*$2^{(n+1)+2}+3^{2(n+1)+1}$
*$2^{n+3}+3^{2n+3}$
*I got stuck here.
How can I prove it using induction?
|
Let $$f(n)=2^{n+2}+3^{2n+1}$$ so that $$f(n+1)=2^{n+3}+3^{2n+3}$$
Then
$$ \begin{align} f(n+1)-f(n) &= 2^{n+3}+3^{2n+3}-2^{n+2}-3^{2n+1} \\
&= 2^{n+2} \left(2-1 \right)+3^{2n+1}\left(3^2 - 1 \right) \\
&= 2^{n+2} + 8\left(3^{2n+1}\right) \\
&= 2^{n+2} + 3^{2n+1} + 7\left(3^{2n+1}\right) \end{align} $$
We assumed that $2^{n+2} + 3^{2n+1} ~|~ 7$ and $7\left(3^{2n+1}\right)$ is obviously divisble by 7. So we're done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1286522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
}
|
Inequality for sides and height of right angle triangle
Someone recently posed the question to me for the above, is c+h or a+b greater, without originally the x and y lengths. I used this method: (mainly pythagorus)
$a^2+b^2=c^2=(x+y)^2=x^2+y^2+2xy$
$a^2=x^2+h^2$ and $b^2=y^2+h^2$
therefore $x^2+h^2+y^2+h^2=x^2+y^2+2xy$
$x^2+y^2+2h^2=x^2+y^2+2xy$
so $2h^2=2xy$ and $$xy=h^2$$
also $Area={ab\over 2}={ch\over 2}$ so $$ab=ch$$
$(a+b)^2=a^2+b^2+2ab$
$(c+h)^2=c^2+h^2+2ch=a^2+b^2+xy+2ab$
therefore $$(a+b)^2+xy=(c+h)^2$$
so $$c+h>a+b$$
I feel like I have made a mistake somewhere, is this incorrectly generalised? And is there an easier way to show the inequality. Also, if this is correct can it be expanded to non right angle triangle, I tried to do this using trig but was pretty much going in circles, thanks!
|
It is not true for all triangles.
From the law of cosines, $c^2 = b^2+ a^2 - 2ab \cos \gamma$, we can see that it is true when the triangle is obtuse.
On the other hand, it is not true for a triangle where all sides are equal.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1286755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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|
How to find differential equation $$\frac{dy}{dx}-8x=2xy^2\quad y=0\,x=1$$
I separated $x$ and $y$.
\begin{align*}
\color{red}{\frac{dy}{y^2}}&=\color{red}{2x+8x dx}\\
\frac{dy}{y^2}&=\color{red}{10x dx}\\
\color{red}{\ln y^2} &= 5x^2\\
y^2&=Ae^{5x^2}
\end{align*}
When I plug $y$ and $x$ in, i get $A=0$.
I think I did it wrong somewhere.
After I fixed it, I get.
$dy/y^2+4=2xdx$
$\ln(y^2+4)=x^2$
$y^2+4=Ae^{x^2}$
After plug in $y$ and $x$
$4=Ae^1$
$A=1.471$
|
$y′=2xy^2+8x⇒y′=x(2y^2+8).$ By separation of variables you find that $$y(x)=2\tan(2(2c_1+x^2))$$ and with the boundary condition $y(1 )=0$ the solution is $$y(x)=2\tan(2(x^2-1))$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1287070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Cannot understand an Integral $$\displaystyle \int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } $$
I had to solve the integral and get it in this form.
My attempt:
$$\int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } $$
$$=\int _{\frac{\pi}{6}}^{ \frac{\pi}{3}} \dfrac{\sin x \cos x }{ \sin x+\cos x }dx $$
Substituting $t=\tan(\frac{x}{2})$,
$$\int_{\tan(\frac{\pi}{12})}^{\tan(\frac{\pi}{6})} \dfrac{2t}{1+t^2}\times\dfrac{1-t^2}{1+t^2}\times\dfrac{2}{1+t^2}dt$$
$$2\int_{2-\sqrt{3}}^{\frac{1}{\sqrt{3}}} \dfrac{2t(1-t^2)}{(1+t^2)^3}dt$$
Substituting $u=1+t^2$, $2t dt=du$, $1-t^2 = 2-u$
$$2\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{(2-u)}{u^3}du$$
$$\displaystyle 4\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{du}{u^3} \displaystyle -2\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{du}{u^2}$$ Could somebody please tell me where I have gone wrong? Also could someone please tell me how to change the limits of the definite integral throughout?
|
Here is an alternate method you could use:
Multiply $\displaystyle\int\frac{\sin x\cos x}{\sin x+\cos x}dx$ on the top and bottom by $\cos x-\sin x$ to get
$\hspace{.6 in}\displaystyle\int\frac{\cos^2x\sin x}{2\cos^2 x-1}dx-\int\frac{\sin^2x\cos x}{1-2\sin^2 x}dx$.
Now substitute $u=\cos x$ in the first integral and $u=\sin x$ in the second integral to get
$\displaystyle\int\frac{u^2}{1-2u^2}du=\frac{1}{2}\int\big(-1+\frac{1}{1-2u^2}\big)du=\frac{1}{2}\bigg[-u+\frac{\sqrt{2}}{4}\ln\bigg|\frac{1+\sqrt{2}u}{1-\sqrt{2}u}\bigg|\bigg]+C$
using partial fractions.
Now you can let $u=\cos x$ in the first term and $u=\sin x$ in the second to get
$\displaystyle\frac{1}{2}\bigg[-u+\frac{\sqrt{2}}{4}\ln\bigg|\frac{1+\sqrt{2}u}{1-\sqrt{2}u}\bigg|\bigg]_{\frac{\sqrt{3}}{2}}^{\frac{1}{2}}-\frac{1}{2}\bigg[-u+\frac{\sqrt{2}}{4}\ln\bigg|\frac{1+\sqrt{2}u}{1-\sqrt{2}u}\bigg|\bigg]_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}$
Notice that substituting $t=\tan\frac{x}{2}$ gives
$\displaystyle\int{\frac{\sin x\cos x}{\sin x+ \cos x} dx=\int\frac{\frac{2t}{1+t^2}\cdot\frac{1-t^2}{1+t^2}}{\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}}}\cdot\frac{2}{1+t^2}dt=\int\frac{4t(1-t^2)}{(1+t^2)^2(1+2t-t^2)}dt$,
and now you can use partial fractions to continue.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1288034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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|
About right identity which is not left identity in a ring Let $S$ be the subset of $M_2(\mathbb{R})$ consisting of all matrices of the form
$\begin{pmatrix}
a & a \\
b & b
\end{pmatrix}$
The matrix $\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}$ is right identity in $S$ if and only if $x+y=1$. Fine, I can see that.
But I cannot see why "If $x+y=1$ , then $\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}$ is not a left identity in $S$".
I have tried that, if $\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}$ is a left inverse then : $\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}\begin{pmatrix}
a & a \\
b & b
\end{pmatrix}=\begin{pmatrix}
x(a+b) & x(a+b) \\
y(a+b) & y(a+b)
\end{pmatrix}=\begin{pmatrix}
a & a \\
b & b
\end{pmatrix}$ in which case we have $x(a+b)=a$ and $y(a+b)=b$. What can i do with $x+y=1$?
|
To prove that $S$ contains no left identity, let an arbitrary
element
$
A
=
\begin{bmatrix}
x & x \\
y & y
\end{bmatrix}
\in S $ be given. Now, either $x = 0$ or $x \ne 0$. If $x = 0$,
then note that
$
\begin{bmatrix}
1 & 1 \\
0 & 0
\end{bmatrix}
\in S
$,
but
\begin{equation*}
A
\begin{bmatrix}
1 & 1 \\
0 & 0
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 \\
y & y
\end{bmatrix}
\begin{bmatrix}
1 & 1 \\
0 & 0
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 \\
y & y
\end{bmatrix}
\ne
\begin{bmatrix}
1 & 1 \\
0 & 0
\end{bmatrix},
\end{equation*}
so $A$ is not a left identity in $S$. If, on the other hand, $x
\ne 0$, then note that
$
\begin{bmatrix}
0 & 0 \\
1 & 1
\end{bmatrix}
\in S
$,
but
\begin{equation*}
A
\begin{bmatrix}
0 & 0 \\
1 & 1
\end{bmatrix}
=
\begin{bmatrix}
x & x \\
y & y
\end{bmatrix}
\begin{bmatrix}
0 & 0 \\
1 & 1
\end{bmatrix}
=
\begin{bmatrix}
x & x \\
y & y
\end{bmatrix}
\ne
\begin{bmatrix}
0 & 0 \\
1 & 1
\end{bmatrix},
\end{equation*}
so $A$ is not a left identity in $S$. Therefore, no element of
$S$ can be a left identity in $S$. That is, $S$ does not contain
a left identity.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1293167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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|
How to find the integral with $\sqrt [ 3 ]{ x } +\sqrt [ 4 ]{ x } $ in the denominator? How to evaluate $$\int { \frac { 1 }{ \sqrt [ 3 ]{ x } +\sqrt [ 4 ]{ x } } } +\frac { \log { (1+\sqrt [ 6 ]{ x } ) } }{ \sqrt [ 3 ]{ x } +\sqrt { x } } dx$$ I'm not being able to make the right substitution.Help please!
|
when you have so many fractional powers, to simplify things try to go for a substitution that can clear up all the powers (usually the LCM of all denominators of various powers will do the job). For example, here let $x=t^{12}$, then you get
\begin{align*}
\int {\frac{1}{\sqrt[3]{x} +\sqrt[4]{x}}} +\frac{\log{(1+\sqrt[6]{x})}}{\sqrt[3]{x}+\sqrt{x}}\, dx & = \int\left[ \frac{1}{t^4+t^3}+ \frac{\log(1+t^2)}{t^4+t^6}\right] \, (12t^{11}) \, dt\\
&= 12 \left[\int \frac{t^8}{1+t} \, dt + \int \frac{t^7 \log(1+t^2)}{1+t^2} \, dt\right]\\
& = 12 \left[\int \frac{t^8-1+1}{1+t} \, dt + \int \frac{t^7 \log(1+t^2)}{1+t^2} \, dt\right]\\
& = 12 \left[\int \frac{1}{1+t} \, dt + \int \frac{t^8-1}{t+1} \, dt + \int \frac{t^7 \log(1+t^2)}{1+t^2} \, dt\right]
\end{align*}
Hopefully now you can solve this. First integral is easy, second is simple factorization, the third is substitution and using integration by parts.
|
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"timestamp": "2023-03-29T00:00:00",
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|
how to parameterize the ellipse $x^2 + xy + 3y^2 = 1$ with $\sin \theta$ and $\cos \theta$ I am trying draw the ellipse $x^2 + xy + 3y^2 = 1$ so I can draw it. Starting from the matrix:
$$ \left[ \begin{array}{cc} 1 & \frac{1}{2} \\ \frac{1}{2} & 3 \end{array}\right]$$
I computed the eigenvalues $2 \pm \frac{1}{2}\sqrt{5}$ and the eigenvectors (not normalized):
$$\left[ \begin{array}{c} x\\ y \end{array}\right]
= \left[ \begin{array}{c} 1\\ 2\pm \sqrt{5} \end{array}\right] $$
So then I tried writing down some combination of the data I generated:
$$ \left[ \begin{array}{c} x(\theta)\\ y(\theta) \end{array}\right]
= \cos \theta \left[ \begin{array}{c} 1\\ 2+ \sqrt{5} \end{array}\right] + \sin \theta \left[ \begin{array}{c} 1\\ 2- \sqrt{5} \end{array}\right]
$$
However, I have a hard time checking the ellipse equation holds true for all $\theta$:
$$x(\theta)^2 + x(\theta)y(\theta) + 3y(\theta)^2 = 1$$
What are the correct functions $x(\theta), y(\theta)$ ?
Following the comments, rescaling the eigenvectors and multiplying the eigenvalues:
$$ \left[ \begin{array}{c} x(\theta)\\ y(\theta) \end{array}\right]
= \frac{2 + \frac{1}{2}\sqrt{5}}{\sqrt{10 + 4 \sqrt{5}}} \left[ \begin{array}{c} 1\\ 2+ \sqrt{5} \end{array}\right]\cos \theta +
\frac{2 - \frac{1}{2}\sqrt{5}}{\sqrt{10 - 4 \sqrt{5}}}\left[ \begin{array}{c} 1\\ 2- \sqrt{5} \end{array}\right]\sin \theta
$$
Is it clear that the ellipse equation is satisfied? I am not sure how to check this.
|
You have $x^2 +xy + 3y^2=(x+1/2 y)^2+11/4 y^2=1$.
You can then take:
$$\begin{array}{lll}
\sin \theta & = & x+1/2 y\\
\cos \theta &= &\sqrt{11}/2 y
\end{array}$$
Which is equivalent to:
$$\begin{array}{lll}
y &= 2/\sqrt{11} \cos \theta\\
x &=\sin \theta - 1/\sqrt{11} \cos \theta
\end{array}$$
to get a parametrization like the one you're looking for.
|
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|
Find $\Big\{ (a,b)\ \Big|\ \big|a\big|+\big|b\big|\ge 2/\sqrt{3}\ \text{ and }\forall x \in\mathbb{R}\ \big|a\sin x + b\sin 2x\big|\le 1\Big\}$
Find all (real) numbers $a $ and $b$ such that $|a| + |b| \ge 2/\sqrt{3} $ and for any $x$ the inequality $|a\sin x + b \sin 2x | \le 1$ holds.
In other words, find the set $Q$ defined as
$$Q = \Big\{\ (a,b)\ \Big| \quad 1. \left|a\right| + \left|b\right| \ge \frac{2}{\sqrt{3}}, \ \text{ and } \ 2. \ \big|\,a\sin x + b \sin 2x \,\big| \le 1 \ \ \ \forall x \in \mathbb{R} \Big\}.$$
This is the problem from one of the recent years Moscow Mathematics Olympiads for 10th or 11th grade. I have no idea how to approach it.
|
using condition and $sin(2x)=2sinx\cdot cosx$ we have for $c\in (0,1]$
$|a+2b\sqrt{1-c^2}|\le \dfrac 1 c \\$
$|a-2b\sqrt{1-c^2}|\le \dfrac 1 c \Rightarrow \\$
$|a|+2|b|\sqrt{1-c^2}\le \dfrac 1 c \\$
$c=\dfrac {\sqrt3} 2 \Rightarrow |a|+|b|\le \dfrac 2 {\sqrt3} \Rightarrow |a|+|b|=\dfrac 2 {\sqrt3} \\ $
so we have
$\dfrac 2 {\sqrt3}-|b|+2|b|\sqrt{1-c^2}\le \dfrac 1 {c}\Leftrightarrow \\$
$|b|(2\sqrt{1-c^2}-1)\le\dfrac 1 c$ so now easily
$c\le \dfrac {\sqrt3} 2 \Rightarrow |b| \le \dfrac 1 {c(2\sqrt{1-c^2}-1)} \\$
$c\ge \dfrac {\sqrt3} 2 \Rightarrow |b| \ge \dfrac 1 {c(2\sqrt{1-c^2}-1)}$
so we have following
$ |b|=lim_{c\rightarrow \frac{\sqrt3} 2} = \dfrac 1 {c(2\sqrt{1-c^2}-1)}=\dfrac 2 {3\sqrt3} \Rightarrow a=\dfrac 4 {3\sqrt3}$
and now we are left to prove the inequality
$|\dfrac {4sinx} {3\sqrt3}+\dfrac {2sin(2x)} {3\sqrt3}|\le 1 \Leftrightarrow \\$
$|2sinx+sin(2x)|\le \dfrac{3\sqrt3} 2 \\$
which follows from AmGm inequality
$|2sinx+sin(2x)|\le \dfrac{3\sqrt3} 2 =2|sinx|\cdot |1+cosx|= \\ \dfrac 2 {\sqrt3} \cdot |\sqrt3 \cdot sinx|\cdot |1+cosx|\le \dfrac 1 {\sqrt3}\cdot \{3sin^2 x+(1+cosx)^2 \}= \\$
$\dfrac 1 {\sqrt3}\cdot \{\dfrac 9 2 -2(cosx-\dfrac 1 2)^2 \}\le \dfrac {3\sqrt3} 2 \\$
equality holds if and only if $x=\dfrac {\pi} 3 +2n\pi,x=\dfrac {-\pi} 3 +2n\pi$
|
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|
Trigonometric equation $\sec(3\theta/2) = -2$ - brain dead Find $\theta$ with $\sec(3\theta/2)=-2$ on the interval $[0, 2\pi]$. I started off with $\cos(3 \theta/2)=-1/2$, thus $3\theta/2 = 2\pi/3$, but I don't know what to do afterwards, the answer should be a huge list of $\theta$s, which I cannot seem to get.
|
Your procedure is correct, except you forgot to include the many other values of $\cos^{-1}(-\frac12)$.
$\cos{x} = -\frac12$ for $x = \frac{2\pi}{3}\pm2\pi n, \frac{4\pi}{3}\pm2\pi n$, where $n$ is an integer.
So you would have:
$$ \frac{3\theta}{2} = \frac{2\pi}{3} \to \theta = \frac{4\pi}{9} $$
$$ \frac{3\theta}{2} = \frac{8\pi}{3} \to \theta = \frac{16\pi}{9} $$
$$ \frac{3\theta}{2} = \frac{14\pi}{3} \to \theta = \frac{28\pi}{9} $$
$$ \frac{3\theta}{2} = \frac{20\pi}{3} \to \theta = \frac{40\pi}{9} $$
etc.
AND:
$$ \frac{3\theta}{2} = \frac{4\pi}{3} \to \theta = \frac{8\pi}{9} $$
$$ \frac{3\theta}{2} = \frac{10\pi}{3} \to \theta = \frac{20\pi}{9} $$
$$ \frac{3\theta}{2} = \frac{16\pi}{3} \to \theta = \frac{32\pi}{9} $$
$$ \frac{3\theta}{2} = \frac{22\pi}{3} \to \theta = \frac{44\pi}{9} $$
etc.
The full list of solutions would be:
$$\theta = \frac{4\pi}{9}\pm\frac{4\pi}{3}$$
$$\theta = \frac{8\pi}{9}\pm\frac{4\pi}{3}$$
|
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|
The lines $x+2y+3=0$ , $x+2y-7=0$ and $2x-y+4=0$ are sides of a square. Equation of the remaining side is? I found out the area between parallel lines as $ \frac{10}{\sqrt{5}} $ and then I used
$ \frac{|\lambda - 4|}{\sqrt{5}} = \frac{10}{\sqrt{5}} $ to get the values as $-6$ and $14$ . I am getting the final equations as $2x-y-6=0$ and $2x-y+14=0$ but this answer is wrong. According to my book the correct equations are $2x-y+6=0$ and $2x-y-14=0$. Please tell me where I am wrong!
|
The first two lines are parallel to each other, and so you are looking for a line parallel to the third one. You have
*
*$x+2y+3=0$
*$x+2y-7=0$
*$2x+(-1)y+4=0$
*$2x+(-1)y+K=0$
where $K$ is an unknown coefficient to complete the square.
The distance between the first and second line (square side) is
$$ d_{12} = \frac{(3)-(-7)}{\sqrt{1^2+2^2}} = 2\sqrt{5} = 4.4721\ldots$$
The same distance should exist between the third and fourth line
$$ d_{34} = \frac{4-K}{\sqrt{2^2+(-1)^2}} = d_{12}$$
$$ K = 4-\sqrt{5} d_{12} = -6$$
So the equation is $$2x-y-6=0$$
this is verified with GeoGebra.
|
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|
Using Gauss elimination to check for linear dependence I have been trying to establish if certain vectors are linearly dependent and have become confused (in many ways). when inputting the vectors into my augmented matrix should they be done as columns or as rows ?
eg if my vectors are [ 4 -1 2 ], [-4 10 2]
I am looking to solve a[ 4 2 2 ] + b[2 3 9] = [0,0,0]
but am unsure how the matrix should be created
$\begin{bmatrix}
4 & -1 & 2 \\
-4 & 10 & 2 \\
\end{bmatrix}$
vs
$\begin{bmatrix}
4 & -4\\
-1 & 10\\
2 & -2
\end{bmatrix}$
using the last method I end up with the following in row reduced echelon form (including the 0's as answers on the right)
$\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 &0
\end{bmatrix}$
so to me this says a = 0, b = 0 , but I don't get what the last row means ?
|
If $ \: \: a \begin{pmatrix}
4 \\
2 \\
2
\end{pmatrix} $ + $ b \begin{pmatrix}
2 \\
3 \\
9
\end{pmatrix} $ = $ \begin{pmatrix}
0 \\
0 \\
0
\end{pmatrix} \: \:$ then $ \: \: \begin{pmatrix}
4a+2b \\
2a+3b\\
2a+9b
\end{pmatrix} $ = $ \begin{pmatrix}
0 \\
0 \\
0
\end{pmatrix}$.
So we have three simultaneous equations,
$ 4a+2b=0 \quad (1)\\ 2a+3b=0 \quad (2)\\ a+9b=0 \quad (3)$
We can now form a coefficient matrix,
Let $ \: A = \begin{pmatrix}
4 & 2 \\
2 & 3 \\
2 & 9
\end{pmatrix} $
Then we put this into reduced row echelon form (RREF) using Gauss-Jordan elimination,
and we get $\; \; \begin{pmatrix}
1 & 0 \\
0 & 1 \\
0 & 0
\end{pmatrix} $.
This tells us that there is a unique solution to the simultaneous equation formed above and that is $ a=0 $ and $ b=0 $.
The reason that the bottom row is zero is because the last of the simultaneous equations can be derived from the other two.
$ (3) = 4*(2) - \frac{5}{2}(1) $.
So it doesn't actually give us any more information.
So if the only solution is $a=0=b$, the vectors $ \begin{pmatrix}
4 \\
2 \\
2
\end{pmatrix} $ and $ \begin{pmatrix}
2 \\
3 \\
9
\end{pmatrix} $ must be linearly independent.
However, it should be clear from the 'top' elements, 4 and 2, that $b=-2a$, but the the 'middle' elements, 2 and 3, need $b=-\frac{2}{3}a$. So the vectors must be independent.
|
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|
The four straight lines given by the equation $12x^2+7xy-12y^2 =0$ and $12x^2+7xy-12y^2-x+7y-1=0$ lie along the side of the? I know these equations are called general equation of second degree and also represent a pair of straight lines. I could extract lines from the equation $$12x^2+7xy-12y^2 =0 $$ (these are $$ 3x+4y=0$$ and $$4x-3y=0$$) but I don't know how to separate lines from $$12x^2+7xy-12y^2-x+7y-1=0.$$ I think we have to find the solution by finding out the difference between all four straight lines but I have trouble separating them. Thank you!
|
In order to avoid the cumbersome calculations, assume that the lines are: $y=m_{1}x+c_{1}$ & $y=m_{2}x+c_{2}$ Now, the quadratic equation of pair of the lines is given as
$$(m_{1}x-y+c_{1})(m_{2}x-y+c_{2})=0 $$$$\implies m_{1}m_{2}x^2-(m_{1}+m_{2})xy+y^2+(m_{1}c_{2}+m_{2}c_{1})x-(c_{1}+c_{2})y+c_{1}c_{2}=0 \tag 1$$ The given equation of pair of lines : $12x^2+7xy-12y^2-x+7y-1=0$ can be written as
$$-x^2-\frac{7}{12}xy+y^2+\frac{1}{12}x-\frac{7}{12}y+\frac{1}{12}=0 \tag 2$$ Now, compare both the above equations (1) & (2), we have $$m_{1}m_{2}=-1 \quad \text{&} \quad m_{1}+m_{2}=\frac{7}{12} \tag 3$$$$ \implies m_{1}-m_{2}=\sqrt{(m_{1}+m_{2})^2-4m_{1}m_{2}}=\sqrt{\left(\frac{7}{12}\right)^2-4(-1)}=\frac{25}{12}\tag 4$$
(Note: $m_{1}$ & $m_{2}$ are unknown hence, their signs will be automatically decided after calculation)
Now, Solving the equations (3) & (4), we get $$m_{1}=\frac{4}{3} \quad \text{&} \quad m_{2}=-\frac{3}{4}$$
Similarly, by comparing the above equations (1) & (2), we have $$c_{1}c_{2}=\frac{1}{12} \quad \text{&} \quad c_{1}+c_{2}=\frac{7}{12} \tag 5$$$$ \implies c_{1}-c_{2}=\sqrt{(c_{1}+c_{2})^2-4c_{1}c_{2}}=\sqrt{\left(\frac{7}{12}\right)^2-4\left(\frac{1}{12}\right)}=\frac{1}{12}\tag 6$$
(Note: $c_{1}$ & $c_{2}$ are unknown hence, their signs will be automatically decided after calculationc)
Now, Solving the equations (5) & (6), we get $$c_{1}=\frac{1}{3} \quad \text{&} \quad c_{2}=\frac{1}{4}$$ Hence, by setting the corresponding values, we get the equations of the lines as follows $$y=\frac{4}{3}x+\frac{1}{3} \quad \text{&} \quad y=-\frac{3}{4}x+\frac{1}{4}$$
$$\implies 4x-3y+1=0 \quad \text{&} \quad 3x+4y-1=0$$
|
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|
How do I find the factorial of a decimal How do I find the following:
$$(0.5)!(-0.5)!$$
Can someone help me step by step here?
|
Factorial of any real number $n$ is defined by Gamma function as follows:
$$\Gamma (n) = (n-1)!$$
$$\quad \Rightarrow ( \dfrac{1}{2} )! ( -\dfrac{1}{2} ) ! = ( \dfrac{3}{2}-1 ) ! ( \dfrac{1}{2}-1 ) ! = \Gamma ( \dfrac {3} {2} ) \Gamma ( \dfrac {1}{2} )$$
It is also known that:
$$\Gamma {(1+z)} = z\Gamma {(z)}$$
$$\quad \Rightarrow \Gamma ( \dfrac {3} {2} ) \Gamma ( \dfrac {1}{2} ) = \dfrac {1}{2} \Gamma ( \dfrac {1}{2} )\Gamma ( \dfrac {1}{2} ) = \dfrac {1}{2} \left( \Gamma ( \dfrac {1}{2} ) \right)^2 $$
Since $\Gamma ( \dfrac {1}{2} ) = \sqrt{\pi}$, then we have:
$$ \quad \Rightarrow \left(\dfrac{1}{2} \right)! \left( -\dfrac{1}{2} \right) ! = \dfrac {1}{2} \left( \Gamma ( \dfrac {1}{2} ) \right)^2 = \dfrac {\pi} {2} $$
|
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|
Finding basis made of uninvertible matrices Let there be transformation $T: \mathbb R_3[X] \rightarrow M_{2 \times 2}(\mathbb R)$, $T(ax^3+bx^2+cx+d)=\left[ \begin{matrix}
a+d & b-2c \\
a+b-2c+d & 2c-b \\
\end{matrix} \right] $
Find a basis of $Im(T)$ made of non-invertible matrices.
So we get that $Im(T)= sp\{\left[ \begin{matrix}
1 & 0 \\
1 & 0 \\
\end{matrix} \right], \left[ \begin{matrix}
0 & 1 \\
1 & -1 \\
\end{matrix} \right], \left[ \begin{matrix}
0 & -2 \\
-2 & 2 \\
\end{matrix} \right],\left[ \begin{matrix}
1 & 0 \\
1 & 0 \\
\end{matrix} \right]\}=sp\{\left[ \begin{matrix}
1 & 0 \\
1 & 0 \\
\end{matrix} \right], \left[ \begin{matrix}
0 & 1 \\
1 & -1 \\
\end{matrix} \right] \}$
And that's also the basis. The first matrix is not invertible, but the second one is. How do I find a basis made of non-invertible matrices?
Thank you for your time!
|
$sp\{\left[ \begin{matrix}
1 & 0 \\
1 & 0 \\
\end{matrix} \right], \left[ \begin{matrix}
0 & 1 \\
1 & -1 \\
\end{matrix} \right] \} =
sp\{\left[ \begin{matrix}
\frac{1}{2} & 0 \\
\frac{1}{2} & 0 \\
\end{matrix} \right], \left[ \begin{matrix}
-\frac{1}{2} & 1 \\
\frac{1}{2} & -1 \\
\end{matrix} \right] \}$
that should do it.
|
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|
How to evaluate $\lim\limits_{n\to\infty}\frac{1}{n}((x+\frac{a}{n})^2+(x+\frac{2a}{n})^2+...+(x+\frac{(n-1)a}{n})^2)$ I don't know how to transform the expression $\frac{1}{n}((x+\frac{a}{n})^2+(x+\frac{2a}{n})^2+...+(x+\frac{(n-1)a}{n})^2)$
The solution, after transformation is
$\frac{n-1}{n}x^2+2\frac{1+...+(n-1)}{n^2}ax+\frac{1^2+...+(n-1)^2}{n^3}a^2$
Thanks for replies.
|
We have $$\lim\limits_{n\to\infty}\frac{1}{n}((x+\frac{a}{n})^2+(x+\frac{2a}{n})^2+...+(x+\frac{(n-1)a}{n})^2)=\lim\limits_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n-1}(x+\frac{k}{n}a)^2$$ and so we have that $$\lim\limits_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n-1}(x+\frac{k}{n}a)^2=\int_{0}^{1}(x+ta)^2 dt=x^2+\frac{a^2}{3}+ax+c$$
|
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|
Integration by Parts Problem: Help in understanding why a part of it equals 0 $$4I= \int_0^{\infty} \frac{4x^3 +\sin(3x)-3\sin x}{x^5} \ \mathrm{d}x $$ $$=\frac{-1}{4} \underbrace{\left[\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} \right]_0^{\infty}}_{=0} +\frac{1}{4} \int_0^{\infty} \frac{12x^2 -3\cos x +3\cos(3x)}{x^4} \ \mathrm{d}x $$
Now, I was told that the brackets gives zero, but not why it is so. Could somebody please help explain this to me? Many thanks!
|
Let's evaluate
$${\left[\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} \right]_0^{\infty}} =$$
$$\lim_{x\to \infty}(\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} )-\lim_{x\to 0}(\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} )$$
take a look on $$\lim_{x\to \infty}(\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} )$$
$sin(3x)$ and $3sin(x)$ are a bounded functions and the deg($4x^3$)< deg($x^4$).
Then you'll find that the value of this limit, when x goes to infinity, is cero.
What about,
$$\lim_{x\to 0}(\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} )$$
It's obvious that the functions $f(x)=4x^3+\sin(3x)- 3 \sin x$ and $g(x)={x^4}$ cab be derived four times so if we apply L'Hopital theorem 4 times we'll find that
$$\lim_{x\to 0}(\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} )=\lim_{x\to 0}(\frac{\ 81sin(3x)- 3 \ sinx}{24} )=0$$
In conclusion $${\left[\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} \right]_0^{\infty}} =0-0=0$$
|
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|
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer
and there are positive integers
$x$ and $y$ such that
$x^2-ny^2 = 1$,
then
$\sqrt{n}$ is irrational.
The proof is in two parts,
each of which
has a one line proof.
Part 1:
Lemma: If
$x^2-ny^2 = 1$,
then there are arbitrarily large integers
$u$ and $v$ such that
$u^2-nv^2 = 1$.
Proof of part 1:
Apply the identity
$(x^2+ny^2)^2-n(2xy)^2
=(x^2-ny^2)^2
$
as many times as needed.
Part 2:
Lemma: If
$x^2-ny^2 = 1$
and
$\sqrt{n} = \frac{a}{b}$
then
$x < b$.
Proof of part 2:
$1
= x^2-ny^2
= x^2-\frac{a^2}{b^2}y^2
= \frac{x^2b^2-y^2a^2}{b^2}
$
or
$b^2
= x^2b^2-y^2a^2
= (xb-ya)(xb+ya)
\ge xb+ya
> xb
$
so
$x < b$.
These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.
Two things to note about
this proof.
First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.
Second,
the key property of
positive integers needed
is that
if $n > 0$
then
$n \ge 1$.
|
Consider the linear application $A:\mathbb{R}^2\to \mathbb{R}^2$ given by $$A=\begin{pmatrix}
-1&2 \\ 1&-1
\end{pmatrix} .$$ $A$ maps $\mathbb{Z}^2$ into itself and $V=\{y=\sqrt 2 x\}$ is an eigenspace relative to the eigenvalue $\sqrt 2-1$. But $A\mid_V$ is a contraction mapping, so $\mathbb{Z}^2\cap V=\emptyset$.
|
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|
prove that the expression $\frac{(3n)!}{(3!)^n}$ is integral for $n \geq 0$ My concept of real no. Is not very clear. Please also tell the logic behind the question.
The expression is true for 19, is it true for all the multiples?
|
You want to prove $\frac{(3n)!}{6^n}$ is an integer. Just use $\frac{3n!}{6^n}=\frac{1\cdot2\cdot3}{6}\frac{4\cdot 5\cdot 6}{6}\frac{7\cdot8\cdot 9}{6}\dots \frac{(3n-2)(3n-1)(3n)}{6}$ and each fraction is an integer since $k(k+1)(k+2)$ is always a multiple of $2$ and of $3$ since three consecutive integers always contain a multiple of $3$ and an odd number. Alternatively because the product of $n$ consecutive numbers is always divisible by $n!$.
I had misunderstood your question as prove $\frac{(3n)!}{n!^3}$ is an integer. Here is a solution to that problem.
Solution 1:
$$\frac{(3n)!}{n!^3}=\frac{1\times2\times\dots \times n}{n!}\frac{(n+1)\times (n+2)\times (n+n)!}{n!}\frac{(2n+1)\times (2n+2)\times \dots (2n+n)}{n!}$$
all three factors are integers since the product of $n$ consecutive integers is divisible by $n!$
Solution $2$:
$$\frac{(3n)!}{n!^3}=\frac{1\times2\times\dots \times n}{n!}\frac{(n+1)\times (n+2)\times (n+n)!}{n!}\frac{(2n+1)\times (2n+2)\times \dots (2n+n)}{n!}=\binom{3n}{n}\binom{2n}{n}\binom{n}{n}$$
Look up binomial coefficient.
Solution $3$:
$$\frac{(3n)!}{n!^3}=\binom{3n}{n,n,n}$$
Look up multinomial coefficient.
|
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|
Find probability of exactly one $6$ in first ten rolls of die, given two $6$s in twenty rolls I am trying to calculate the probability that, when rolling a fair die twenty times, I roll exactly one $6$ in the first ten rolls, given that I roll two $6$s in the twenty rolls.
My thoughts
Let $A = \{\text {Exactly one 6 in first ten rolls of a die} \}$ and $B = \{\text {Exactly two 6s in twenty rolls of a die} \}.$
Then I want to find
$$P[A\mid B] = \frac{P[A \cap B]}{P[B]}.$$
By the binomial distribution formula, we get that
$$P[B] = {20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}.$$
Furthermore I think that $P[A \cap B]$ is equal to the probability of rolling exactly one $6$ in ten rolls and then rolling exactly one $6$ in another set of ten rolls. That is,
$$P[A \cap B] = \left[{10 \choose 1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(\frac{5}{6}\right)^9\right]^2.$$
Am I correct in thinking this?
If so, then it follows that the required probability is
$$P[A \mid B] = \frac{\left[{10 \choose 1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(\frac{5}{6}\right)^9\right]^2}{{20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}},$$
which, I know, can be simplified further!
|
I took a different approach to the question. Suppose B. There are three ways to get two 6's in twenty rolls:
*
*$B_1$: Both 6's come in the first 10 rolls. There are $\begin{pmatrix} 10 \\ 2\end{pmatrix} = 45$ ways for this to happen.
*$B_2$: One 6 comes in the first 10 rolls, and the second comes in the next 10 rolls. There are $\begin{pmatrix} 10 \\ 1\end{pmatrix} \begin{pmatrix} 10 \\ 1\end{pmatrix} = 100$ ways for this to happen.
*$B_3$: Both 6's come in the second lot of 10 rolls. There are $\begin{pmatrix} 10 \\ 2\end{pmatrix} = 45$ ways for this to happen.
Now
$$P(A|B) = P(B_2|B_1 \cup B_2 \cup B_3) = \frac{100}{45+100+45} = \frac{10}{19}.$$
|
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|
Modular arithmetic , calculate $54^{2013}\pmod{280}$.
How do you calculate: $54^{2013}\pmod{280}$?
I'm stuck because $\gcd(54,280)$ is not $1$. Thanks.
|
As $280=2^3\cdot5\cdot7$
let us start with $54^{2013-3}\pmod{\dfrac{280}{2^3}}$ i.e., $54^{2010}\pmod{35}$
Now $54\equiv-1\pmod5\implies54^2\equiv(-1)^2\equiv1$
and $54\equiv-2\pmod7\implies54^3\equiv(-2)^3\equiv-1,54^6\equiv(-1)^2\equiv1$
$\implies54^6\equiv1\pmod{35}$
As $2010\equiv0\pmod6,54^{2010}\equiv54^0\pmod{35}\equiv1$
As $a\equiv b\pmod m\implies a\cdot c\equiv b\cdot c\pmod{m\cdot c}\ \ \ \ (1)$
$54^{2010}\cdot54^3\equiv1\cdot54^3\pmod{35\cdot54^3}$
As $280|35\cdot54^3,54^{2010+3}\equiv54^3\pmod{280}$
Again as $54=2\cdot3^3,54^3=\cdots=2^33^9$
and $3^3\equiv-8\pmod{35},8=2^3;3^9\equiv-(2^3)^3\equiv-2^9$
Now $2^7\equiv-12\pmod{35}\implies2^9\equiv-12\cdot2^2\equiv-13$
$\implies3^9\equiv13\pmod{35}$
Using $(1),54^3=2^3\cdot3^9\equiv2^3\cdot13\pmod{2^3\cdot35}\equiv104$
|
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|
Integer partitioning Suppose we have an integer $n$. I we want to partition the integer in the form of $2$ and $3$ only; i.e., $10$ can be partitioned in the form $2+2+2+2+2$ and $2+2+3+3$.
So, given an integer, how to calculate the total number of ways of doing such partitions and how many $2$'s and $3$'s are there in each of the partitions?
|
For even $n$, the number of $3$ in each partition is even. So, let $2k$ be the largest number of $3$ in the partition of even $n$, i.e.
$$3\cdot 2k\le n\lt 3(2k+2)\Rightarrow k\le \frac{n}{6}\lt k+1\Rightarrow k=\left\lfloor\frac n6\right\rfloor.$$
Hence, the number of partitions of $\color{red}{\text{even}\ n}$ is $\color{red}{\left\lfloor\frac{n}{6}\right\rfloor+1}$ (note that the +1 comes from the case $k=0$).
For odd $n$, the number of $3$ in each partition is odd. So, let $2k-1$ be the largest number of $3$ in the partition of odd $n$, i.e.
$$3(2k-1)\le n\lt 3(2k+1)\Rightarrow k\le \frac{n+3}{6}\lt k+1\Rightarrow k=\left\lfloor\frac{n+3}{6}\right\rfloor.$$
Hence, the number of partitions of $\color{red}{\text{odd}\ n}$ is $\color{red}{\left\lfloor\frac{n+3}{6}\right\rfloor}$.
|
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|
What two numbers when multiplied gives $-25$ and when added, $-10$? Factors of $25$ are; $1, 5, 25$
$$5 \times 5 = 25$$
$$5 + 5 = 10$$
$$-5 \times 5 = -25$$
$$-5 + 5 = 0$$
How can I solve this?
Thanks
|
Solve: $$\begin{align}
xy &=-25 \\
x+y &= -10
\end{align}
$$
In particular, substituting $y=-10-x = -(10+x)$ in the first equation, you get $$x^2+10x -25 = 0 $$
which is an equation you should know how to handle. The two solutions are $x=-5+\sqrt{50} = 5(\sqrt{2}-1)$ and $x=-5(\sqrt{2}+1)$, from which you get respectively $y=-5(\sqrt{2}+1)$ and $y=5(\sqrt{2}-1)$.
|
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|
solving difficult complex number proving if $z= x+iy$ where $y \neq 0$ and $1+z^2 \neq 0$, show that the number $w= z/(1+z^2)$ is real only if $|z|=1$
solution :
$$1+z^2 = 1+ x^2 - y^2 +2xyi$$
$$(1+ x^2 - y^2 +2xyi)(1+ x^2 - y^2 -2xyi)=(1+ x^2 - y^2)^2 - (2xyi)^2$$
real component
$$(1+ x^2 - y^2)x - yi(2xyi) = x + x^3 + xy^2$$
imaginary component
$$-2yx^2 i +yi + x^2 yi - y^3 i=0i$$
$$-2yx^2 +y + x^2 y - y^3 =0$$
...
can't solve this question
|
$$\dfrac Z{1+Z^2}=\dfrac{x+iy}{1+x^2-y^2+2xyi}$$
$$=\dfrac{(x+iy)(1+x^2-y^2-2xyi)}{(1+x^2-y^2)^2+(2xy)^2}$$
We need $y(1+x^2-y^2)-x(2xy)=0\implies y(1+x^2-y^2-2x^2)=0\iff x^2+y^2=1$
|
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|
Limit of $\dfrac{(1+4^x)}{(1+3^x)}$? I don't remember how to find the limit in this case. I take $x$ towards $+\infty$.
$\lim\limits_{x\to \infty} \dfrac{1+4^x}{1+3^x}$
I do not know where to start. I would instinctively say that $1$ can't be right because $4^x$ goes faster than $3^x$ and thus one would move towards infinity, but this is apparently not the case....
|
$$\lim_{x\rightarrow \infty}\dfrac{(1+4^x)}{(1+3^x)}=$$
$$\lim_{x\rightarrow \infty}\frac{1}{1+3^x}+\lim_{x\rightarrow \infty}\frac{4^x}{1+3^x}=$$
$$\frac{1}{\lim_{x\rightarrow \infty} 1+3^x}+\lim_{x\rightarrow \infty}\frac{4^x}{1+3^x}=$$
$$\frac{1}{\infty}+\lim_{x\rightarrow \infty}\frac{4^x}{1+3^x}=$$
$$0+\lim_{x\rightarrow \infty}\frac{4^x}{1+3^x}=$$
$$\lim_{x\rightarrow \infty}\frac{\left(\frac{4}{3}\right)^x}{3^{-x}+1}=$$
$$\lim_{x\rightarrow \infty}\frac{\left(\frac{4}{3}\right)^x}{0+1}=$$
$$\lim_{x\rightarrow \infty}\frac{\left(\frac{4}{3}\right)^x}{1}=$$
$$\lim_{x\rightarrow \infty}\left(\frac{4}{3}\right)^x=$$
$$\left(\frac{4}{3}\right)^{\lim_{x\rightarrow \infty}x}=\left(\frac{4}{3}\right)^{\infty}=\infty$$
|
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|
Show if $\sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}}$ is absolute convergent or divergent Show if $\sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}}$ is absolute convergent or divergent
First i subbed numbers in
$$\lim_{n \to \infty} \frac{(-1)^n}{1+\sqrt{n}} = \frac{-1}{1+\sqrt{1}} + \frac{1}{1+\sqrt{2}} - \frac{-1}{1+\sqrt{3}}$$
So it's divergent
1 LIMIT
$$\lim_{n \to \infty} \frac{1}{1+\sqrt{n}}=0 \quad \text{hence divergent} $$
2 $a_{n}$ and $a_{n+1}$
$$ \frac{1}{1+\sqrt{n}}>\frac{1}{1+\sqrt{n+1}} $$
$$ \frac{1}{1+\sqrt{n}}-\frac{1}{1+\sqrt{n+1}}> 0 $$
hence increasing
or
$$ \frac{1}{1+\sqrt{n}}-\frac{1}{1+\sqrt{n+1}} = \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n+1}} $$
$$ \therefore \sqrt{n+1}-\sqrt{n}> 0$$ hence decreasing
$ \sum_{n=1}^{\infty} \left\lvert \frac{-1^n}{1+\sqrt{n}} \right\rvert = \sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}} $
$$ \sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}} = \frac{1}{1+\sqrt{1}} + \frac{1}{1+\sqrt{2}} + \frac{1}{1+\sqrt{3}} $$
if this is convergent then the whole thing is absolutely convergent but i don't know how to prove this
UPDATE
I got something out for the second part
|
Let $a_n=\frac{(-1)^n}{1+\sqrt{n}}$.
$$ a_{2n}+a_{2n+1} = \frac{1}{1+\sqrt{2n}}-\frac{1}{1+\sqrt{2n+1}} = \frac{\sqrt{2n+1}-\sqrt{2n}}{(1+\sqrt{2n})(1+\sqrt{2n+1})} = \sqrt{2n} \frac{\sqrt{1+\frac{1}{2n}}-1}{(1+\sqrt{2n})(1+\sqrt{2n+1})} \sim \frac{n^{-\frac{3}{2}}}{4\sqrt{2} } = O\left(\frac{1}{n^{\frac{3}{2}}}\right)$$
Therefore the serie $\sum (a_{2n}+a_{2n+1})$ is convergent and so is the serie $\sum a_n$.
Moreover, $|a_n| \sim \frac{1}{\sqrt{n}} $ so your serie is not absolutely convergent.
|
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|
How to solve this inequality, with the hypothesis more complicated than the conclusion? Given $x,y,z \in \mathbb{R}$ and $x,y,z>2,$ I want to show that if,
$$\frac{1}{x^2-4}+\frac{1}{y^2-4}+\frac{1}{z^2-4} = \frac{1}{7}$$
then,
$$\frac{1}{x+2} + \frac{1}{y+2} + \frac{1}{z+2} \leq \frac{3}{7}.$$
I follow the solution here: http://artofproblemsolving.com/community/c6h514107_inequality_by_poru_loh
but I don't know how to alter it to fit this problem?!
|
Here is an adaptation of pi37 answer in your given link.
Note
$$\sum_{cyc}\dfrac{x^2+25}{x^2-4}=\sum_{cyc}\dfrac{x^2-4+29}{x^2-4}=3+\dfrac{29}{7}=\dfrac{50}{7}$$
and use AM-GM $x^2+25\ge 10x$.so
$$\sum_{cyc}\dfrac{x^2+25}{x^2-4}\ge\dfrac{10x}{x^2-4}\Longrightarrow\sum_{cyc}\dfrac{x}{x^2-4}\le \dfrac{5}{7}$$
and note
$$\sum_{cyc}\dfrac{x}{x^2-4}-\sum_{cyc}\dfrac{2}{x^2-4}=\sum_{cyc}\dfrac{x-2}{x^2-4}=\sum_{cyc}\dfrac{1}{x+2}$$
so
$$\sum_{cyc}\dfrac{1}{x+2}\le\dfrac{5}{7}-\dfrac{2}{7}=\dfrac{3}{7}$$
|
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|
How to find out the greater number from $15^{1/20}$ and $20^{1/15}$? I have two numbers $15^{\frac{1}{20}}$ & $20^{\frac{1}{15}}$.
How to find out the greater number out of above two?
I am in 12th grade. Thanks for help!
|
Well raise both numbers to the power of $20$
That is
$$\large{(15^\frac{1}{20})^{20} = 15^\frac{20}{20} = 15}$$
Now $$\large{(20^\frac{1}{15})^{20} = 20^\frac{20}{15} = 20^\frac{4}{3} = 20^{1.333..}}$$
which is greater ? $\large{15}$ or $\large{20^{1.333...}}$
Clearly , it is $\large{20^{1.333..}}$ because $\large{20^{1.333} > 20^1 > 15}$ and so this means that $\large{20^\frac{1}{15} >15^\frac{1}{20}}$
|
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Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is I got this task two days ago, quite difficult for me, since I have not done applications of Vieta's formulas and Bezout's Theorem for a while. If can someone solve this and add exactly how I am supposed to use these two theorem's on this task, I would be thankful.
Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is equal to?
|
\begin{align*}
x^4+ax^2+b & = x^2\color{blue}{(x^2+4x+6)}-4x\color{blue}{(x^2+4x+6)}+\\
&(a+10)\color{blue}{(x^2+4x+6)}-\color{red}{(4a+16)(x)+(b-6a-60)}
\end{align*}
For $x^2+4x+6$ to divide the given polynomial.We need $4a+16=0$ and $b-6a-60=0$. Thus $a=-4$ and $b=36$.
|
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|
Solving a complex number inequality involving absolute values. Here is the relevant paragraph (from "Complex numbers from A to Z" by Titu Andreescu and Dorin Andrica) :
Original question : How does $\left | 1+z \right |=t$ imply $\left | 1-z+z^2 \right |=\sqrt{\left | 7-2t^2 \right |}$?
(I checked for $z=i$ , it seems it is wrong ...)
EDIT: It seems it is indeed wrong. So , how can I prove the inequality?(perhaps even the lower and upper bounds need to be changed)
|
Since $|z|$, we can write $$ z= \cos\theta+ \sin\theta$$
Now we can derive useful properties for such $z$:
$1+z = 1 + \cos\theta + i\sin\theta = 1 + 2\cos^2\frac{\theta}{2} - 1 + 2i\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2\cos\frac{\theta}{2}(\cos\frac{\theta}{2} + i \sin\frac{\theta}{2})$
Thus, $|1+z| = 2|\cos\frac{\theta}{2}|$. Besides,
$$1 + z^2-z = 1 + \cos2\theta+ i\sin2\theta - (\cos\theta + i\sin\theta) = $$
$$2\cos\theta(\cos\theta + i\sin\theta) - (\cos\theta + i\sin\theta) = (2\cos\theta-1)(\cos\theta + i\sin\theta) $$
Then,
$$| 1 -z + z^2| = |2\cos\theta -1| = \left|4\cos^2\frac{\theta}{2} -3 \right| $$
If you define $x=\cos\frac{\theta}{2}$, with $x \in [-1,1]$, then we have:
$$|1+ z|+ |1 - z + z^2| = 2|x| + |4x^2 -3| = f(x)$$
Whose graph is easy to draw. Just divide the interval of $x$ in pieces such that the expressions inside the modules have constant sign. To make it easier, note that $f(x)$ is even. Therefore, you can do:
$$x \in \left[0,\frac{\sqrt3}{2}\right] \Rightarrow f(x) = -4x^2 +2x +3$$
$$x \in \left[\frac{\sqrt3}{2},1 \right] \Rightarrow f(x) = 4x^2 +2x -3$$
Which gives $f(-\frac{1}{4})=\frac{13}{4}$ for the maximum and $f(\frac{\sqrt3}{2}) = \sqrt3$ for the minimum.
|
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|
Divisibility of a polynomial by another polynomial I have this question:
Find all numbers $n\geq 1$ for which the polynomial $x^{n+1}+x^n+1$ is divisible by $x^2-x+1$. How do I even begin?
So far I have that $x^{n+1}+x^n+1 = x^{n-1}(x^2-x+1)+2x^n-x^{n-1}+1,$ and so the problem is equivalent to finding $n$ such that $2x^n-x^{n-1}+1$ is divisible by $x^2-x+1.$
A solution that I found goes as follows (but I don't understand it!):
Assume that $x^n+1=(x^m-x+1)Q(x).$ The polynomial $x^m-x+1$ has a real root in the interval $(0,1)$ but $x^n+1$ has no positive real roots. So, no such pairs $m,n$ exist.
Any help?
|
Hint: The roots of $x^2-x+1=0$ are $e^{i\pi/3}$ and $e^{-i\pi/3}$. If we show that one (and therefore the other) cannot be a root of $x^{n+1}+x^n+1$, then we will know that $x^2-x+1$ cannot divide $x^{n+1}+x^n+1$.
There are $3$ cases to examine: (i) $n$ is of the form $3k+2$; (ii) $n$ is of the form $3k+1$; and (iii) $n$ is of the form $3k$.
|
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Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$
My attempt is
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)+2-\sqrt{3}$$
$$x^3=4+\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)$$
then what I will do??
|
$$x^3=(r+s)^3=r^3+3r^2s+3rs^2+s^3=r^3+s^3+3rs(r+s)=4+3x,$$
because $r^3+s^3=4$ and $rs=1$.
|
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|
Maximum value of trigonometric expression If
$r=3+\tan c \tan a, \quad
q=5+\tan b \tan c, \quad
p=7+\tan a \tan b$
Provided $a,b,c$ are positive and
$a+b+c=\dfrac{\pi}2$
Find the maximum value of
$\sqrt p + \sqrt q + \sqrt r$ .
|
Let $x = \tan a \tan b, \; y = \tan b \tan c, \; z = \tan c \tan a$, then $x, y, z > 0$ and $x+y+z=1$.
We need to now maximize $\sqrt{3+x}+\sqrt{5+y}+\sqrt{7+z}$. As $\sqrt t$ is concave, using Karamata's inequality and $(3+x, 5+y, 7+z) \succ (3+x+y+z, 5, 7) = (4, 5, 7)$, we have $\sqrt{3+x}+\sqrt{5+y}+\sqrt{7+z} \le 2+\sqrt5 + \sqrt7$.
Equality is never achieved, but one can get as close to the value when $b \ll a \to 0^+, c \to \frac{\pi}2^-$, so we do not have a maximum but a supremum.
|
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About the "Cantor volume" of the $n$-dimensional unit ball A simple derivation for the Lebesgue measure of the euclidean unit ball in $\mathbb{R}^n$ follows from computing
$$ \int_{\mathbb{R}^n}e^{-\|x\|^2}\,dx $$
in two different ways. See, for instance, Keith Ball, An Elementary Introduction to Modern Convex Geometry, page $5$. Now I was wondering about the following slightly unusual variation:
Let $X_1,\ldots,X_n$ independent random variables with the Cantor
distribution.
What is the probability that $X_1^2+\ldots+X_n^2\leq 1$?
I bet this can be tackled by exploiting the fact that the cumulants of the Cantor distribution are given by:
$$ \kappa_{2n}=\frac{2^{2n-1}(2^{2n}-1)\,B_{2n}}{n (3^{2n}-1)},\tag{CM}$$
but how to prove $(\mathrm{CM})$? - This has been answered, but the main question is still open.
|
HINT:
For $\mu(x)$ the Cantor measure supported on the Cantor set $\subset [0,1]$ we have the change of variable formula:
$$\int f(x)\, d\mu(x) = \frac{1}{2} \int f(1/3 x)\, d\mu(x) + \frac{1}{2} \int f(1/3 x + 2/3)\, d \mu(x)$$
analogous to $\int_0^1 f(x)\, dx =\frac{1}{2} \int_0^1 f(1/2 x)\, d x + \frac{1}{2} \int_0^1 f(1/2 x + 1/2)\, d x $
$\bf{Added:}$
It's easy to see that the first moment $E(X) = \int x \, d \mu(x)= \frac{1}{2}$, and this can be obtained readily from the above formula for $f(x) = x$.
Consider now the central moment generating function
$$F(t)\colon =E[e^{t(X-\frac{1}{2})}] = \int e^ {t(x-\frac{1}{2})} \, d\mu(x) $$
From the above equality for $f_t(x) = e^{t(x-\frac{1}{2})}$ we get
$$\int e^ {t(x-\frac{1}{2})} \, d\mu(x) = \frac{1}{2}\left( \int e^ {t(\frac{x}{3}-\frac{1}{2})} \, d\mu(x) + \int e^ {t(\frac{x}{3}+\frac{2}{3}-\frac{1}{2})} \, d\mu(x) \right) $$
Now we notice that
\begin{eqnarray}
t\,(\frac{x}{3}-\frac{1}{2})= \frac{t}{3}(x - \frac{1}{2}) - \frac{t}{3}\\
t\,(\frac{x}{3}+\frac{2}{3}-\frac{1}{2})= \frac{t}{3}(x - \frac{1}{2}) + \frac{t}{3}
\end{eqnarray}
Therefore we get the equality
$$F(t) = \frac{e^{\frac{t}{3}} + e^{-\frac{t}{3}}}{2} \cdot F(\frac{t}{3})
$$
$\bf{Added:}$ Rewrite the above equality as
$$F(3t) = \frac{e^t + e^{-t}}{2} F(t)$$
Let $G(t) = \log F(t)$. From the above we get
$$G(3 t) - G(t) = \log ( \frac{e^t + e^{-t}}{2})$$
$\bf{Added:}$
Some (moment) calculations:
$$\int x d \mu(x) = \frac{1}{2} \left( \ \int (\frac{1}{3} x + \frac{1}{3} x + \frac{2}{3}) d\mu(x) \right )$$
implies $\int x d \mu(x) = \frac{1}{2}$ as expected.
Let's apply the same formula for $f(x) = (x-\frac{1}{2})^n$. We have
\begin{eqnarray}
\int (x-\frac{1}{2})^n d \mu(x) = \frac{1}{2}\left( \int ( \frac{x}{3} - \frac{1}{2})^n + ( \frac{x}{3} + \frac{2}{3}- \frac{1}{2})^n d\mu(x) \right ) = \\
=\frac{1}{2\cdot 3^n}\left( \int ( x - \frac{1}{2}-1)^n + ( x-\frac{1}{2} + 1)^n d\mu(x) \right )
\end{eqnarray}
that is
$$M_n = \frac{1}{3^n}\sum_{k \ge 0} \binom{n}{2k} M_{n-2k}$$
which is basically a formula from above $F(3t) = \frac{e^t + e^{-t}}{2} F(t)$.
We get from here $m_2 = \frac{1}{8}$, $m_4 = \frac{7}{320}$, etc. Note that the formula provided in Wikipedia is for the cumulants, not the central moments, as $\kappa_4 = \frac{1}{40}$.
|
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|
Derivative and graph mismatch Using the implicit function $(x^2+y^2-1)^3=x^2y^3$ it can be shown that $y'=\frac{2xy^3-6x(x^2+y^2-1)^2}{6y(x^2+y^2-1)^2-3x^2y^2}$ but when I evaluate it for the point (1,0) I get $y'(1,0)=\frac{0}{0}$ even though the slope of the tangent line is 2 at that point.
Any ideas?
Garth
|
Instead $(x^2+y^2-1)^3=x^2y^3$ we can take $x^2+y^2-1=x^{2/3}y$ near of $(1,0)$, hence
\begin{align*}
\left(y-\frac{1}{2}x^{2/3}\right)^2&=1-x^2+\frac{x^{4/3}}{4}\\
y&=-\left(1-x^2+\frac{x^{4/3}}{4}\right)^{1/2}+\frac{x^{2/3}}{2}\\
y'&=-\frac{1}{2}\left(1-x^2+\frac{x^{4/3}}{4}\right)^{-1/2}\left(-2x+\frac{x^{1/3}}{3}\right)+\frac{1}{3}x^{-1/3}
\end{align*}
Then
\begin{align*}
y'(1)&=-\frac{1}{2}\left(1-1+\frac{1}{4}\right)^{-1/2}\left(-2+\frac{1}{3}\right)+\frac{1}{3}\\
&=-\frac{1}{2}(2)\left(-\frac{5}{3}\right)+\frac{1}{3}\\
&=2
\end{align*}
|
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|
How to evaluate $\lim_{n\to\infty} a_n$ If $a_1=1$ and $a_{n+1}=\frac{4+3a_n}{3+2a_n}$,$n\geq1$, then how to prove that $a_{n+2}>a_{n+1}$ and if $a_n$ has a limit as ${n\to\infty}$ then how to evaluate $\lim_{n\to\infty} a_n$ ?
|
Consider the function $f(x) = \frac{4+3x}{3+2x} = \frac32 - \frac{1}{2(3+2x)}$. It is easy to check.
*
*$f(x)$ is a strictly increasing function from $[0,\infty)$ to $[0,\infty)$.
*$\sqrt{2}$ is a fixed point for $f(x)$.
So start from any number $a \in (0,\sqrt{2})$, if we construct a sequence $a_n$ by
$$a_{n} = \begin{cases} a, & n = 1\\ f(a_{n-1}), & n > 1\end{cases}$$
We will have
$$a_1 < \sqrt{2}
\implies a_2 = f(a_1) < f(\sqrt{2}) = \sqrt{2}
\implies a_3 = f(a_2) < f(\sqrt{2}) = \sqrt{2}
\implies \cdots$$
This means all $a_n < \sqrt{2}$. By setting $a = 1$, we will recover the
sequence $a_n$ we have. Notice
$$a_1 = 1 \quad\text{ and }\quad a_2 = f(a_1) = f(1) = \frac75 > a_1$$
The strictly increasing property of $f(x)$ leads to
$$a_1 < a_2 \implies a_2 = f(a_1) < f(a_2) = a_3 \implies a_3 = f(a_2) < f(a_3) = a_4 \implies \cdots$$
This means the sequence we have is monotonic increasing.
Combine these two observation, $a_n$ is a monotonic increasing sequence, bounded from above by $\sqrt{2}$. So it converges to some limit $a_\infty \le \sqrt{2}$. Since $f(x)$ is continuous, we have
$$f(a_\infty) = f(\lim_{n\to\infty} a_n) = \lim_{n\to\infty} f(a_n) = \lim_{n\to\infty} a_{n+1} = a_\infty$$
So $a_\infty$ is a fixed point of $f(x)$. Solving the equation $f(x) = x$ for $x \in [1,\sqrt{2}]$ leads to $a_\infty = \sqrt{2}$.
For this particular problem, there is actually a way to compute $a_n$ explicitly.
Rewrite $a_n$ as $\frac{p_n}{q_n}$ for some $p_n, q_n$ to be determined. We can rewrite the recurrence relation as
$$a_{n+1} = \frac{3a_n + 4}{2 a_n + 3} \quad\iff\quad
\frac{p_{n+1}}{q_{n+1}} = \frac{3p_n + 4 q_n}{2 p_n + 3q_n}\tag{*1}$$
If we scale $p_n, q_n$ so that they satisfies following matrix equation
$$\begin{bmatrix}p_{n+1}\\ q_{n+1}\end{bmatrix}
= \begin{bmatrix}3 & 4\\ 2 & 3\end{bmatrix}
\begin{bmatrix}p_{n}\\ q_{n}\end{bmatrix}$$
It will solve the RHS of $(*1)$. What this means is if we define two sequences
$p_n, q_n$ by
$$
\begin{bmatrix}p_{n}\\ q_{n}\end{bmatrix} =
\begin{bmatrix}3 & 4\\ 2 & 3\end{bmatrix}^{n-1}
\begin{bmatrix}1\\1\end{bmatrix}\quad\text{ for } n \ge 1$$
Our $a_n$ will be equal the quotient of these two pair of sequences. Diagonalizing the matrix and with a little bit of algebra, one find
$$a_n = \sqrt{2}\left[\frac{
(\sqrt{2}+1)^{2n-1} - (\sqrt{2}-1)^{2n-1}
}{
(\sqrt{2}+1)^{2n-1} + (\sqrt{2}-1)^{2n-1}
}\right]$$
Using this representation, it is immediately clear $\lim_{n\to\infty} a_n = \sqrt{2}$.
|
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|
Prove or disprove that $8c+1$ is square number.
Let $a,b,c$ be positive integers, with $a-b$ prime, and $$3c^2=c(a+b)+ab.$$ Prove or disprove that $8c+1$ is square number.
|
Yes, that's true! Here we go: Write $b=a-p$. Then, if you write the expression as a quadratic equation in $c$, you get $3c^2-(2a-p)c+ap-a^2=0.$ Now, by the quadratic formula, see that since we want $c$ to be integer, we must have $(2a-p)^2-12(ap-a^2)=x^2$ for some, say positive, integer $x$. Rewriting in a better way, you get $(4a-2p)^2=x^2+3p^2$, which turns to be $(4a-2p-x)(4a-2p+x)=3p^2.$ Now, since $p$ is prime, there are a few different cases we should look at:
1) $4a-2p-x=1, 4a-2p+x=3p^2$: Isolating $2a-p$ and $x$, you get $x=\frac{3p^2-1}{2}, 2a-p=\frac{3p^2+1}{4}.$ Plugging in these two in quadratic formula, you get $c=\frac{9p^2-1}{24}$ or $c=\frac{-3p^2+3}{24}$, where first one cannot be an integer since numerator is not divisible by $3$, and the latter is not positive. So, there is no solution from this case.
2) $4a-2p-x=3, 4a-2p+x=p^2$. Similarly, you get $2a-p=\frac{p^2+3}{4}, x=\frac{p^2-3}{2}$. Plug in these, and eliminate the possible negative root, you'll get $c=\frac{p^2-1}{8}$, which tells that $8c+1=p^2.$
3) $4a-2p-x=p, 4a-2p+x=3p$. From here, you get $x=p, 2a-p=\frac{p}{2}$, which forces $p$ to be $2$, so you get $a=\frac{3}{2}$, hence no solution from this case too.
Note that these are all possible cases, since $4a-2p-x<4a-2p+x.$
|
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|
Find $x+y+z$ from the equation In the equation
$$\dfrac{37}{13}=2+\dfrac{1}{x+\dfrac{1}{y+\dfrac 1z}}$$
find the value of $x+y+z$
How can I know the value of three variables while there is only one equation?
|
$$\frac{26+11}{13}=2+\frac{1}{\frac{13}{11}}=2+\frac{1}{1+\frac{1}{\frac{11}{2}}}=2+\frac{1}{1+\frac{1}{5+\frac{1}{2}}}$$
|
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|
Partial fraction of $\frac 1{x^6+1}$ Can someone please help me find the partial fraction of $$1\over{x^6+1}$$
?
I know the general method of how to find the partial fraction of functions but this seems a special case to me..
|
\begin{align}
\frac 1{x^6+1}&=\frac 1{(x^2)^3+1}\\
&=\frac 1{((x^2)1+)((x^2)^2-(x^2)+1)} \\
&= \frac 1{(x^2+1)(x^4-x^2+1)} \\
&= \frac {Ax+B}{x^2+1}+\frac{Cx+D}{x^4-x^2+1}
\end{align}
So $$1=(x^4-x^2+1)(Ax+B)+(x^2+1)(Cx+D)$$
At $x=0$, the equation becomes $$1=B+D$$
At this point, equating the coefficients is the only way to go.
|
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|
solve $x^2 \equiv 24 \pmod {60}$ I need to solve $x^2 \equiv 24 \pmod {60}$
My first question which confuses me a lot -
isn't a (24 here) has to be coprime to n (60)???
most of the theorems requests that.
what i tried -
$ 60 = 2^2 * 3 * 5$
So I need to solve $x^2 \equiv 24$ modulo each one of $2^2, 3, 5$ so i get -
$x^2 \equiv 0 \pmod 4$
$x^2 \equiv 0 \pmod 3$
$x^2 \equiv 4 \pmod 5$
so if Im correct I have 5 equations -
$x \equiv 0 \pmod 4$
$x \equiv 2 \pmod 4$
$x \equiv 0 \pmod 3$
$x \equiv 2 \pmod 5$
$x \equiv 3 \pmod 5$
Now my questions are - am I correct until now.
and second is, how to solve this? I know to use the Chinese reminder theorem
but what confuses me here is that I have more then one equation modulo the same number.
any help will be appreciated .
|
You were correct.
$$x^2\equiv 24\pmod{\! 60}\iff \begin{cases}x^2\equiv 24\equiv 0\pmod{\! 3}\\ x^2\equiv 24\equiv 0\pmod{\! 4}\\ x^2\equiv 24\equiv 4\pmod{\! 5}\end{cases}$$
$$\iff \begin{cases}x\equiv 0\pmod{\! 3}\\ x\equiv 0\pmod{\! 2}\\ x\equiv \pm 2\pmod{\! 5}\end{cases}$$
If and only if at least one of the two cases holds:
$1)$ $\ x\equiv 0\pmod{\! 6},\ x\equiv 2\pmod{\! 5}$
$2)$ $\ x\equiv 0\pmod{\! 6},\ x\equiv -2\pmod{\! 5}$
You can use Chinese Remainder theorem as follows (when I create new variables, they're integers):
$$x\equiv 0\pmod{\! 6}\iff x=6k$$
$$1)\ \ \ x\equiv 2\pmod{\! 5}\iff \color{#00F}6k\equiv \color{#00F}1k\equiv 2\pmod{\! 5}$$
$x=6(5n+2)=30n+12$.
$$2)\ \ \ x\equiv 3\pmod{\! 5}\iff \color{#00F}6k\equiv \color{#00F}1k\equiv 3\pmod{\! 5}$$
$x=6(5n+3)=30n+18$.
Another way you can use CRT (which is basically just finding an $x$ that works in $[0,30)$):
$1)\ \ \ (x\equiv 0\equiv 12\pmod{\! 6}$ and $x\equiv 2\equiv 12\pmod{\! 5})\iff x\equiv 12\pmod{\! 6\cdot 5},$
because (since $(6,5)=1$):
$$6,5\mid x-12\iff 6\cdot 5\mid x-12$$
Using this, in case $2)$ in the same way you find that $18$ works ($18\equiv 0\pmod{\! 6}$ and $18\equiv 3\pmod{\! 5}$).
So you have the congruence holds iff $x=30m\pm 12$ for some $m\in\Bbb Z$.
|
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|
Help with indefinite integration I am learning indefinite integration, yet am having problems understanding and recognizing where to substitute what. a good trick is to attempt convert algebraic expressions into trigonometric and vice versa. However, despite doing so, I am unable to solve the integral. For example, in attempting to integrate:
$$\sqrt\frac{1-\sqrt x}{1+ \sqrt x}$$
I substituted $x = \cos^2 t$ in order to convert it to a trigonometric equation. However, I am still unable to solve the integral.
|
Let
\begin{align}
I = \int \sqrt{\frac{1 - \sqrt{x}}{1+\sqrt{x}}} \, dx
\end{align}
and make the substitution $x = t^{2}$ to obtain
\begin{align}
I = 2 \, \int \sqrt{\frac{1-t}{1+t}} \, t \, dt.
\end{align}
Now let $t = \cos(2\theta)$ to obtain
\begin{align}
I &= -4 \, \int \sqrt{\frac{1- \cos(2\theta)}{1+\cos(2\theta)}} \, \, \cos(2\theta) \, \sin(2\theta) \, d\theta \\
&= -2 \, \int \sqrt{\frac{2 \sin^{2}\theta}{2 \cos^{2}\theta}} \, \sin(4\theta) \, d\theta \\
&= -2 \, \int \tan(\theta) \, \sin(4\theta) \, d\theta = 2 \theta - 2 \sin(2\theta) + \frac{\sin(4\theta)}{2}.
\end{align}
Reviewing the substitutions it is evident that $2\theta = \cos^{-1}(\sqrt{x})$ and that
\begin{align}
I = \cos^{-1}(\sqrt{x}) - 2 \, \sqrt{1-x} + \sqrt{x(1-x)} + c_{0}
\end{align}
|
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|
Finding the last digit of $7^n$, $n\ge 1$. I have noticed a cycle of 7,9,3,1. Meaning: $7^1\equiv 7\pmod {10}, 7^2\equiv 9\pmod {10}, 7^3\equiv 3\pmod {10},7^4\equiv 1\pmod {10}, 7^5\equiv 7\pmod {10}$ and so on. Therefore, if $n=4k+1$ the last digit is 7, If $n=4k+2$, the last digit is 9, If $n=4k+3$, the last digit is 3 and if $n=4k$, the last digit is 1. I don't really know how to get to a final, coherent answer. That is as far as I could. I guess I am required to present a specific function the outputs the exact last digit. I would really appreciate your help.
|
Take $n=4m+k$ where $0\leq k\leq 3$, then
$$7^{4m+k}=7^{4m}\cdot 7^k=(7^4)^m\cdot 7^k\equiv 1^m 7^k \equiv 7^k (mod 10).$$
|
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|
Interesting variant of binomial distribution. Suppose i had $n$ Bernoulli trials with $X_{i}=1$ if the $i$th trial is a success and $X_{i}=-1$ if it is a failure each with probability $\frac{1}{2}$.
Then the difference between the number of successes and failures can be represented by the random variable $Y=|\sum_{i=1}^{n} X_{i}|$ what is the best way to calculate $\mathbb{E}(Y)$?
One thing i was thinking is $|\# \mathrm{Successes}-\# \mathrm{Failures}|=n-2\#\mathrm{successes}$ and so $\mathbb{E}(Y)=\mathbb{E}(2|\frac{n}{2}-\#\mathrm{Successes}|)$ which we could relate to the variance of $X=\sum_{i=1}^{n} X_{i}'$? where $X_{i}'$ is an indicator random variable equal to $1$ if $X_{i}$ is a success and $0$ otherwise. Am i along the right lines?
|
Let $S_{2n}$ be the number of successes in $2n$ trials. Then there are $2n - S_{2n}$ failures, with a difference of $|S_{2n} - (2n - S_{2n})| = |2S_{2n} - 2n|$. Now let $0 <k \leq 2n$ be an even number. Then
\begin{align*}
\mathbb{P}(|2S_{2n} - 2n| = k) &= \mathbb{P}(2S_{2n} - 2n = k) + \mathbb{P}(2S_{2n} - 2n = -k) \\
&= \mathbb{P}\left(S_{2n} = n + \frac{k}{2}\right) + \mathbb{P}\left(S_{2n} = n - \frac{k}{2}\right) \\
&= \frac{1}{2^{2n}} \left(\binom{2n}{n + \frac{k}{2}} + \binom{2n}{n - \frac{k}{2}}\right) \\
&= \frac{1}{2^{2n-1}}\binom{2n}{n + \frac{k}{2}}.
\end{align*}
If $k$ is odd, then it is easy to check that $\mathbb{P}(|2S_{2n} - 2n| = k) = 0$. We have
\begin{align*}
\mathbb{E}(Y_{2n}) &= \sum_{k=2}^{2n}k\,\mathbb{P}(|2S_{2n} - 2n| = k)\\
&= \frac{1}{2^{2n-2}}\sum_{k=1}^n k \, \binom{2n}{n + k} \\
&= \frac{n+1}{2^{2n-1}}\binom{2n}{n+1} \\
&= \frac{n+1}{2^{2n-1}} \cdot \frac{(2n)!}{(n+1)!(n-1)!} \\
&= \frac{n+1}{2^{2n-1}} \cdot \frac{(2n)!}{n!n!} \cdot \frac{n}{n+1} \\
&= \boxed{\frac{2n}{2^{2n}} \binom{2n}{n}}.
\end{align*}
Similarly, let $S_{2n+1}$ be the number of successes in $2n+1$ trials, for a difference of $|S_{2n+1} - (2n+1 - S_{2n+1})| = |2S_{2n+1} - 2n - 1|$. Let $1 \leq k \leq 2n+1$ be an odd number in the form $k = 2k' + 1$. Then
\begin{align*}
\mathbb{P}(|2S_{2n+1} - 2n - 1| = k) &= \mathbb{P}(2S_{2n+1} - 2n - 1 = k) + \mathbb{P}(2S_{2n+1} - 2n - 1 = -k) \\
&= \mathbb{P}\left(S_{2n+1} = n + \frac{k+1}{2}\right) + \mathbb{P}\left(S_{2n+1} = n + \frac{1-k}{2}\right) \\
&= \frac{1}{2^{2n+1}} \left(\binom{2n+1}{n + \frac{k+1}{2}} + \binom{2n+1}{n + \frac{1-k}{2}}\right) \\
&= \frac{1}{2^{2n}}\binom{2n+1}{n + \frac{k+1}{2}}.
\end{align*}
Writing $k = 2k' + 1$, the probability becomes
$$\mathbb{P}(|2S_{2n} - 2n - 1| = k) = \frac{1}{2^{2n}}\binom{2n+1}{n + k' + 1}.$$
Even differences are not possible, so we have
\begin{align*}
E(Y_{2n+1}) &= \sum_{k=1}^{2n+1} k \, \mathbb{P} (|2S_{2n+1} - 2n - 1| = k) \\
&= \frac{1}{2^{2n}}\sum_{k'=0}^n (2k'+1) \,\binom{2n+1}{n + k' + 1} \\
&= \frac{n+1}{2^{2n}}\binom{2n+1}{n+1} \\
&= \frac{n+1}{2^{2n}}\cdot \frac{2n+1}{n+1} \cdot \frac{2n!}{n!n!} \\
&= \boxed{\frac{2n+1}{2^{2n}} \binom{2n}{n}}.
\end{align*}
Stirling's approximation states that
$$n! \sim \sqrt{2\pi n}n^n e^{-n},$$
so
$$\binom{2n}{n} \sim \frac{2^{2n}}{\sqrt{\pi n}}.$$
This gives us
$$E(Y_{2n}) \sim \frac{2n}{\sqrt{\pi n}} = \frac{2}{\sqrt{\pi}} \sqrt{n}.$$
Thus the expected difference is $\boxed{\mathcal{O}(\sqrt{n})}.$
EDIT 1: My previous answer didn't have the right conditional expectation, so I just computed general probabilities and used the definition of expectation.
EDIT 2: Added asymptotics.
|
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|
Find remainder when $777^{777}$ is divided by $16$
Find remainder when $777^{777}$ is divided by $16$.
$777=48\times 16+9$. Then $777\equiv 9 \pmod{16}$.
Also by Fermat's theorem, $777^{16-1}\equiv 1 \pmod{16}$ i.e $777^{15}\equiv 1 \pmod{16}$.
Also $777=51\times 15+4$. Therefore,
$777^{777}=777^{51\times 15+4}={(777^{15})}^{51}\cdot777^4\equiv 1^{15}\cdot 9^4 \pmod{16}$ leading to $ 81\cdot81 \pmod{16} \equiv 1 \pmod{16}$.
But answer given for this question is $9$. Please suggest.
|
Hint: $9^2~=~81~=~5\cdot16+1$.
|
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|
Solving $\sin x=2\sin(2\pi/3-x)$ How can I solve the equation: $$\sin x=2\sin\left(\frac{2\pi}{3}-x\right)$$
Without using the formula:
$$\sin(a-b)=\sin a \cos b-\sin b \cos a$$?
Thanks.
|
Consider the next picture
By Sine Law the equality $\frac{\sin x}{2} = \frac{\sin\left(2\pi/3-x\right)}{1}$ holds, then by using the Cosine Law we have
\begin{align*}
c^2&=5-4\cos\frac{\pi}{3}\\
&=3\\
c&=\sqrt{3}
\end{align*}
Now, from Sine Law, it follows.
\begin{align*}
\sin x&=\frac{2\sin \pi/3}{\sqrt{3}}\\
\sin x&=1\\
\implies x&=\frac{\pi}{2}
\end{align*}
|
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|
Sum of solutions of this exponential equations How to solve this :
$$x^{3-\log_{10}(x/3)}=900$$
I tried log on both sides and got nothing with exponent of $x$ and $3$.
|
Go ahead and take the $\log_{10}$ on both sides:
$$3\log_{10}(x)-\log_{10}(x)^2+\log_{10}(x)\log_{10}(3)=\log_{10}(900).$$
Now solve the quadratic.
Let $y=\log_{10}(x).$ Then this quadratic is
$$y^2-(\log_{10}(3)+3)y+\log_{10}(900)=0.$$
Applying the quadratic formula, we get
$$y=\frac{3+\log_{10}(3)\pm\sqrt{\log_{10}(3)^2+6\log_{10}(3)+9-4\log_{10}(900)}}{2}$$
$$=\frac{3+\log_{10}(3)\pm\sqrt{\log_{10}(3)^2+6\log_{10}(3)+9-8\log_{10}(3)-8}}{2}$$
$$=\frac{3+\log_{10}(3)\pm\sqrt{\log_{10}(3)^2-2\log_{10}(3)+1}}{2}$$
$$=\frac{3+\log_{10}(3)\pm(\log_{10}(3)-1)}{2}$$
$$=\cases{1+\log_{10}(3)=\log_{10}(30)\\2}$$
Thus, $x=30$ or $x=100$.
|
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|
Prove this inequality $\frac{1}{1+a}+\frac{2}{1+a+b}<\sqrt{\frac{1}{a}+\frac{1}{b}}$
Let $a,b>0$ show that
$$\dfrac{1}{1+a}+\dfrac{2}{1+a+b}<\sqrt{\dfrac{1}{a}+\dfrac{1}{b}}$$
It suffices to show that
$$\dfrac{(3a+b+3)^2}{((1+a)(1+a+b))^2}<\dfrac{a+b}{ab}$$
or
$$(a+b)[(1+a)(1+a+b)]^2>ab(3a+b+3)^2$$
this idea can't solve it to me,are we aware of an elementary way of proving that? Thanks in advance.
|
if op go ahead , he will get from his last step :LHS-RHS$=(4a^4+4a^2+b^3-6a^2b)+(a^2b^2-3ab+b+a)+a^2b^3+ab^3+3a^3b^2+a^2b^2+ab^2+2b^2+3a^4b+a^3b+a^5+6a^3 >0 \iff$
$(4a^4+4a^2+b^3-6a^2b) \ge 0 \cap (a^2b^2-3ab+b+a)\ge 0$
|
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|
$(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz$? The question given is
Show that $(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz$.
What I tried is suppose $a=(y+z-x),\ b=(z+x-y)$ and $c=(x+y-z)$ and then noted that $a+b+c=x+y+z$. So the question statement reduced to $(a+b+c)^3-(a^3+b^3+c^3)$. Then I tried to invoke the identity $(a^3+b^3+c^3-3abc)=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$ by adding and subtracting $3abc$ in the question statement. After doing all this when I substituted back the values of $a,b$ and $c$, I ended up with the initial question statement.
Any hints will be appreciated.
|
Observe that,
$\begin{align}(x+y+z)^3-(x+y-z)^3&=2z\left((x+y+z)^2+(x+y-z)^2)+(x+y)^2-z^2\right)\\&=2z\left(3(x+y)^2+z^2\right)\tag{1}\end{align}$
and,
$\begin{align}(x-y-z)^3-(x-y+z)^3&=-2z\left((x-y-z)^2+(x-y+z)^2)+(x-y)^2-z^2\right)\\&=2z\left(3(x-y)^2+z^2\right)\tag{2}\end{align}$
Therefore,
$(x+y+z)^3+(x-y-z)^3-(x-y+z)^3-(x+y-z)^3\\=\bigl((x+y+z)^3-(x+y-z)^3\bigr)+\bigl((x-y-z)^3-(x-y+z)^3\bigr)\\=2z\left(3(x+y)^2+z^2\right)-2z\left(3(x-y)^2+z^2\right)\\=2z(3\cdot 4xy)\\=24xy$
|
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|
Hints on solving $y'=\frac{y}{3x-y^2}$
$$y'=\frac{y}{3x-y^2}$$
My attempt:
$$\frac{dy}{dx}=\frac{y}{3x-y^2}$$
$$dy\cdot(3x-y^2)=dx\cdot y$$
$$dy\cdot3x-dy\cdot y^2=dx\cdot y$$
Any direction?
I need hints please $\color{red}{not}$ a full answer
|
Following the hint given by Chinny84, we write the equation as $$\dfrac{\mathrm{d}x}{\mathrm{d}y} = \frac{3x-y^2}{y} = \frac{3x}{y} -y$$
So $$\frac{\mathrm{d}x}{\mathrm{d}y} - \frac{3x}{y} = -y$$ which is a linear first order differential equation in $x$.
Spoiler: Answer below.
Using an integrating factor $$I = \exp\left({\int \frac{-3}{y}}\, \mathrm{d}y \right) = \frac{1}{y^3}.$$
We get that $$\frac{x}{y^3}= \int \frac{1}{y^3}\cdot-y \, \mathrm{d}y = \frac{1}{y} + \mathrm{c}.$$
Multiplying through by $y^3$ yields
$$x = \mathrm{c}y^3 + y^2$$
|
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|
Unusual result to the addition
Question: Prove that
(666... to n digits)^2 + (888... to n digits)=(444... to 2n digits)
My way: I just proved the given equation for three values of n and written at the bottom.
"Since the equation satisfies for n=1, 2, and 3, the equation is true and hence proved."
Also I am seeing a regular pattern in (6666...to n digits)^2
$666^2=443556,6666^2=44435556,66666^2=4444355556$
So the pattern is first there are (n-1) $4's$ the one $3$ then (n-1) 5's the one 6.
Is there a general way to solve this question?
|
Question: Prove that
$(666\dots \text{to $n$ digits})^2 + (888\dots \text{to $n$
> digits})=(444\dots \text{to $2n$ digits})$
By dividing by $4$, this is equivalent to
\begin{align}
(333\dots \text{to $n$ digits})^2
+ (222\dots \text{to $n$ digits})
&=(111\dots \text{to $2n$ digits})
\end{align}
Let $x=111\dots \text{to $n$ digits}$. Then
\begin{align}
(3x)^2+2x&=x\cdot 10^n+x
\\
9x^2+2x&=x\cdot 10^n+x
\\
x(9x+2)&=x\cdot 10^n+x
\\
x(9x+1+1)&=x\cdot 10^n+x
\\
x(10^n+1)&=x\cdot 10^n+x.
\end{align}
|
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|
Find the sum$\pmod{1000}$
Find $$1\cdot 2 - 2\cdot 3 + 3\cdot 4 - \cdots + 2015 \cdot 2016 \pmod{1000}$$
I first tried factoring,
$$2(1 - 3 + 6 - 10 + \cdots + 2015 \cdot 1008)$$
I know that $\pmod{1000}$ is the last three digits.
$$= 2 - 6 + 12 - 20 + 30 - 42 + \cdots + 240$$
that is too complicated. Just hints please!
|
The given series is: $$\sum_{i=1} ^{1008} (2i-1)(2i)-\sum_{i=1} ^{1007}(2i)(2i+1)$$ which is $$4\times\sum_{i=1} ^{1008} i^2-2\times \sum_{i=1} ^{1008} i-4\times \sum_{i=1} ^{1007} i^2+2\times \sum_{i=1} ^{1007} i$$ which, by cancelling out the common terms becomes, $$4\times 1008^2-2\times 1008=2\times 1008\times (2\times 1008-1)=2\times 1008\times 2015$$ which $\pmod {1000}$ is $$2\times 8\times 15=240.$$
|
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|
Can we obtain $f(y+x)=y+f(x)$ from $f(x^2+f(x)^2+x)=f(x)^2+x^2+f(x)$?
$\mathbb Z^+$ is the set of positive integers. Find all functions $f:\mathbb{Z}^+\rightarrow \mathbb{Z}^+$ such that
$$f(m^2+f(n))=f(m)^2+n\quad(\clubsuit)$$
Let $P(x,y)$ be the assertion: $f(x^2+f(y))=f(x)^2+y \; \forall x,y \in \mathbb{Z}^+.$
$P(x,x)$ gives us $f(x^2+f(x))=f(x)^2+x$.
$P(x,x^2+f(x))$ gives us $f(x^2+f(x)^2+x)=f(x)^2+x^2+f(x)$.
Can we obtain $f(y+x)=y+f(x)$ from $f(x^2+f(x)^2+x)=f(x)^2+x^2+f(x)$ ?
|
Given
$$
f: \mathbb{Z} \rightarrow \mathbb{Z} \wedge f(m^2 + f(n)) = f(m)^2 + n.
$$
For a fixed $m$ and $n=x$ we get
$$
f(m^2 + f(x)) = f(m)^2 + x.
$$
In general we can write
$$
f(x) = \sum_k a_k x^k.
$$
Whence
$$
f'(m^2 + f(x)) f'(x) = 1.
$$
Thus $f'(x) = 1$.
So we obtain
$$
f(x) = a + x.
$$
Putting it back we get
$$
a + [ m^2 + a + n ] = a^2 + 2 a m + m^2 + n \Rightarrow 2a = a^2 + 2 a m.
$$
Whence
$$
a = 0.
$$
So the general solutions is
$$
f(x) = x.
$$
|
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|
Evaluating a function at a point where $x =$ matrix. Given $A=\left(
\begin{array} {lcr}
1 & -1\\
2 & 3
\end{array}
\right)$
and $f(x)=x^2-3x+3$ calculate $f(A)$.
I tried to consider the constant $3$ as $3$ times the identity matrix ($3I$) but the answer is wrong. Appreciate any ideas.
|
$$f(A) = A^2 - 3A + 3I$$
But you know that $$A^2 = \begin{pmatrix}
-1 & -4 \\ 8 & 7
\end{pmatrix}$$
So, $$f(A) = \begin{pmatrix}
-1 & -4 \\ 8 & 7
\end{pmatrix} + \begin{pmatrix}
-3 & 3 \\ -6 & -9
\end{pmatrix} + \begin{pmatrix}
3 & 0 \\ 0 & 3
\end{pmatrix}$$
Simplifying leads to $$\bbox[10px, border: 2px solid lightblue]{f(A) = \begin{pmatrix}
-1 & -1 \\ 2 & 1
\end{pmatrix}}$$
|
{
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|
Find 4 numbers which create a ratio Find four numbers which create ratio if its known that sum of first and last is equal to 14, sum of middle two is equal to 11 and sum of squares of all numbers is equal to 221
I got only that sum of product first with last and product of two middle is equal to 48
I have written this question by phone so please sorry if I did mistake
I hope I explain my point clear
|
If we interpret the phrase "four numbers which create ratio" as "four numbers that make a proportion", i.e. "the ratio of the first two numbers equals the ratio of the last two numbers," then we have four equations in the four variables $a,b,c,d$:
$$\frac ab=\frac cd$$
$$a+d=14$$
$$b+c=11$$
$$a^2+b^2+c^2+d^2=221$$
From the middle two we get $d=14-a$ and $c=11-b$. Substituting those into the first equation we get
$$\frac ab=\frac {11-b}{14-a}$$
$$a(14-a)=b(11-b)$$
$$a^2-14a=b^2-11b$$
Substituting the expressions for $c$ and $d$ into the last of the four original equations,
$$a^2+b^2+(11-b)^2+(14-a)^2=221$$
$$a^2+b^2+121-22b+b^2+196-28a+a^2=221$$
$$2(a^2-14a)+2(b^2-11b)+96=0$$
$$2(a^2-14a)+2(a^2-14a)+4\cdot 24=0$$
$$a^2-14a+24=0$$
$$(a-12)(a-2)=0$$
$$a=12 \quad\text{or}\quad a=2$$
If we take $a=12$ then we get
$$12^2-14\cdot 12=b^2-11b$$
$$b^2-11b+24=0$$
$$(b-8)(b-3)=0$$
$$b=8 \quad\text{or}\quad b=3$$
The first gives us the numbers $12,8,3,2$ in order; the second gives us $12,3,8,2$, the same four numbers in a different order.
Similarly, if we had taken $a=2$ we still would have gotten $b=8$ or $b=3$, giving us the answers $2,8,3,12$ or $2,3,8,12$, the very same four numbers in different orders.
Our final answer is the four numbers $12,8,3,2$, in four different possible orders.
|
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|
According to Stewart Calculus Early Transcendentals 5th Edition on page 140, in example 5, how does he simplify this problem? In Stewart's Calculus: Early Transcendentals 5th Edition on page 140, in example 5, how does
$$\lim\limits_{x \to \infty} \frac{\dfrac{1}{x}}{\dfrac{\sqrt{x^2 + 1} + x}{x}}$$
simplify to
$$\lim\limits_{x \to \infty} \frac{\dfrac{1}{x}}{\sqrt{1 + \dfrac{1}{x^2}} + 1}$$
I understand how he simplifies $\frac{x}{x}$ to 1 but how does he simplify $\frac{\sqrt{x^2 + 1}}{x}$?
|
Hint: Observe that, for $x>0$,
$$\large\sqrt{x^2+1\strut}=\sqrt{x^2(1+\tfrac{1}{x^2})}=\sqrt{x^2\strut} \;\cdot\;\sqrt{1+\tfrac{1}{x^2}}=x\;\cdot\;\sqrt{1+\tfrac{1}{x^2}}$$
|
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|
Calculus - Solving limits with square roots I am having trouble understanding how to solve this limit by rationalizing. I have the problem correct (I used Wolfram Alpha of course), but I still don't understand how it is completed. I was trying to solve this by multiplying both the numerator and the denominator by $\sqrt{x^2+11}+6$, but I am stuck. Is that the way to solve this? If so, could you walk me through it so I can solve others?
Problem:
|
$$\begin{align}
\lim_{x\to 5}\frac{\sqrt{x^2+11}-6}{x-5}
&= \lim_{x\to 5}\frac{\left(\sqrt{x^2+11}-6\right)
\left(\sqrt{x^2+11}+6\right)}{(x-5)\left(\sqrt{x^2+11}+6\right)} \\[2ex]
&= \lim_{x\to 5}\frac{\left(\sqrt{x^2+11}\right)^2-6^2}
{(x-5)\left(\sqrt{x^2+11}+6\right)} \\[2ex]
&= \lim_{x\to 5}\frac{x^2+11-36}{(x-5)\left(\sqrt{x^2+11}+6\right)} \\[2ex]
&= \lim_{x\to 5}\frac{x^2-25}{(x-5)\left(\sqrt{x^2+11}+6\right)} \\[2ex]
&= \lim_{x\to 5}\frac{(x-5)(x+5)}{(x-5)\left(\sqrt{x^2+11}+6\right)} \\[2ex]
&= \lim_{x\to 5}\frac{x+5}{\sqrt{x^2+11}+6} \\[2ex]
&= \frac{5+5}{\sqrt{5^2+11}+6} \\[2ex]
&= \frac{10}{12} \\[2ex]
&= \frac{5}{6}
\end{align}$$
|
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|
Hypergeometric distribution exercise! A store has $20$ guitars in stock but 3 are defective. Claire buys $5$ guitars from this lot.
(a) Find the probability that Claire bought $2$ defective guitars.
I use $N=20,n=5, k = 3,x=2$ where $N$ is the total sample space, $n$ is the number of trials, $k$ is the number of defectives and $x$ is the number of guitars bought that are defective
Hence I got $h(2;20,5,3) = \frac{{3 \choose 2}{17 \choose 3}}{20 \choose 5} = 0.1316$
Is my answer correct? I think I may be missing something.
|
We assume that Claire's selection method was such that all collections of $5$ guitars were equally likely to be bought.
There are $\binom{20}{5}$ ways to choose $5$ guitars from $20$.
There are $\binom{3}{2}$ ways to choose $2$ defectives from $3$. For each of these ways, there are $\binom{17}{3}$ ways to choose $3$ non-defectives to accompany them, for a total of $\binom{3}{2}\binom{17}{3}$.
So your expression for the probability is right.
Now let us compute. The numerator is
$3\cdot \frac{(17)(16)(15)}{6}$. This is $2040$. The denominator is
$\frac{(20)(19)(18)(17)(16)}{120}$, which is $15504$. Divide. Your answer is correct to $4$ significant figures.
A simplified fractional form is $\frac{5}{38}$.
|
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|
Divisibility question Prove:
(A) sum of two squares of two odd integers cannot be a perfect square
(B) the product of four consecutive integers is $1$ less than a perfect square
For (A) I let the two odd integers be $2a + 1$ and $2b + 1$ for any integers $a$ and $b$. After completing the expansion for sum of their squares , I could not establish the link to it not being a perfect t square
For (B) I got stuck at expanding $(a)(a + 1)(a+2) (a + 3)$
Could someone help please ?
Thanks
|
A)
\begin{align*}
&\,(2a+1)^2+(2b+1)^2=(4a^2+4a+1)+(4b^2+4b+1)=4(a^2+b^2)+4(a+b)+2\\
=&\,2[2(a^2+b^2)+2(a+b)+1].
\end{align*}
Since $2(a^2+b^2)$ and $2(a+b)$ are both even, the expression between the brackets is odd because of the $+1$ term. Now, the double of an odd number can never be a perfect square. (Try proving this last statement.)
B)
\begin{align*}
&\,a(a+1)(a+2)(a+3)=a(a+3)[(a+1)(a+2)]=a(a+3)[a^2+3a+2]\\
=&\,a(a+3)[a(a+3)+2]=a^2(a+3)^2+2a(a+3)=[a^2(a+3)^2+2a(a+3)+1]-1\\
=&\,[a(a+3)+1]^2-1.
\end{align*}
|
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|
Integrating the given function involving trigonometric functions Find $\int \csc^{p/3}x \sec^{q/3} x dx $
Given - $(p,q \in I^{+} )$ and $(p+q=12)$
I tried to substitute $q = 12-p$ in the integral but didn't find anything satisfactory.
|
$$
\int \csc ^{p/3}(x) \sec^{q/3}(x)dx = \int \frac{1}{\sin^{p/3}(x)\cos^{4-\frac{p}{3}}(x)}dx = \int \frac{1}{\cos^4 x \cdot \tan^{p/3}{x}}dx = \int \frac{\sec^4 x}{\tan ^{p/3}x}dx
$$
Let $u=\tan x \implies du = \sec^2 x dx \implies$
$$
\int\frac{(1+u^2)du}{u^{p/3}} = \frac{3}{3-p}u^{\frac{3-p}{3}} + \frac{3}{9-p}u^{\frac{9-p}{3}} + C = \frac{3}{3-p}\tan^{\frac{3-p}{3}}x + \frac{3}{9-p}\tan^{\frac{9-p}{3}}x + C
$$
|
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|
Integral which must be solved using integration by parts I have to solve this problem using integration by parts. I am new to integration by parts and was hoping someone can help me.
$$\int\frac{x^3}{(x^2+2)^2} dx$$
Here is what I have so far:
$$\int udv = uv-\int vdu $$
$$u=x^2+2$$ Therefore, $$xdx=\frac{du}{2}$$
$$dv=x^3$$
Therefor, $$v=3x^2$$
|
Hint:
$\int\frac{x^3}{(x^2+2)^2}dx$
Write $x^3asx^2x$
$\int\frac{x^2x}{(x^2+2)^2}dx$
*
*add and substract 2 in numerator
$\int\frac{[(x^2+2)-2]x}{(x^2+2)^2}dx$
*
*separate it as two integrals
$\int\frac{(x^2+2)x}{(x^2+2)^2}dx-\int\frac{2x}{(x^2+2)^2}dx$
$=>\int\frac{x}{(x^2+2)}dx-\int\frac{2x}{(x^2+2)^2}dx$
formulae :(1) $\int\frac{f{'}(x)}{f(x)}dx=log|f(x)|$
formulae:(2)$int\frac{f{'}(x)}{f^{2}(x)}dx=-\frac{1}{f(x)}$
-take $f(x) = x^2+2$ for the first part of integral and before that multiply and divide with 2 for using formulae(1)
*
*take $f(x) = x^2+2$for the second integral using formulae ( 2)
Then answer is
$=\frac{1}{2}log|x^2+2|+\frac{1}{x^2+2}$
|
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|
How prove $\lim x_n = \sqrt{\frac{a}{b}}$ if $b \neq 0$ for $x_1=c$ and $x_{n+1}=\frac{x_n^2 + \frac{a}{b}}{2x_n}$? Let $x_1=c$ and $x_{n+1}=\frac{x_n^2 + \frac{a}{b}}{2x_n}$. How prove $\lim x_n = \sqrt{\frac{a}{b}}$ if $b \neq 0$ ?
|
For convenience, let $s:=\sqrt{\dfrac ab}$.
Then,
$$\frac{x_{n+1}-s}{x_{n+1}+s}=\frac{\dfrac{x_n^2+s^2}{2x_n}-s}{\dfrac{x_n^2+s^2}{2x_n}+s}=\left(\frac{x_n-s}{x_n+s}\right)^2,$$
and by recurrence
$$\frac{x_n-s}{x_n+s}=\left(\frac{c-s}{c+s}\right)^{2^n}.$$
This gives us the explicit formula
$$\color{green}{x_n=s\frac{1+r^{2^n}}{1-r^{2^n}}},$$
where $$r:=\frac{c-s}{c+s}.$$
It (quickly) converges to $s$ when $$|r|=\left|\frac{c-s}{c+s}\right|<1,$$
which is true for any
$$c>0.$$
Addendum for numericians:
Choose the integer $k$ such that $2^k\le s^2<2^{k+1}$ ($k$ is the rank of the most significant bit in $s^2$) and let $$c=\frac1{\sqrt[4]2}\cdot2^{k/2}\text{ or }c=\sqrt[4]2\cdot2^{(k-1)/2}$$ depending on the parity of $k$. Then
$$\frac 1{\sqrt[4]2}\le\frac cs<\sqrt[4]2.$$
This makes an excellent starting approximation, as at worse $r=0.086427\cdots=2^{-3.5323\cdots}$ and the iterates give at least $3.5,7,28,56\cdots$ exact bits, which are enough for the single and double precision floating-point representations.
|
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|
Proving the integral series $\int _0^1\left(1-x^2\right)^n\,dx=\frac{2}{3}\cdot \frac{4}{5}\cdot\ldots\cdot \frac{2n}{2n+1}$
We have the series $\left(I_n\right)_{n\ge 1\:}$ where $$I_n=\int _0^1\left(1-x^2\right)^n\,dx.$$
Prove that $$I_n=\frac{2}{3}\cdot \frac{4}{5}\cdot\ldots\cdot \frac{2n}{2n+1}.$$
I tried to integrate that function for $n=1,2,3$ to see whether there's a pattern (recurrence relation), but I just can't figure out how to write the pattern and how to prove that statement. Must I use induction?
|
As Byron Schmuland has pointed out a recursive pattern was developed here. The process in this solution will be a connection to the Beta function.
Consider the integral
\begin{align}
I_{n} = \int_{0}^{1} \left(1-x^{2}\right)^{n} \, dx.
\end{align}
Let $t = x^{2}$ to obtain
\begin{align}
I_{n} &= \frac{1}{2} \, \int_{0}^{1} t^{-\frac{1}{2}} \, (1-t)^{n} \, dt \\
&= \frac{1}{2} \, B\left(\frac{1}{2}, n + 1 \right) \\
&= \frac{\Gamma\left(\frac{3}{2}\right) \, \Gamma\left(n + 1\right)}{\Gamma\left(n + \frac{3}{2}\right)} = \frac{2}{3}\cdot \frac{4}{5}\cdot\ldots\cdot \frac{2n}{2n+1}.
\end{align}
|
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|
$a_1,a_2,...,a_n$ are positive real numbers, their product is equal to $1$, show: $\sum_{i=1}^n a_i^{\frac 1 i} \geq \frac{n+1}2$ it says to use the weighted AM-GM to solve it, because the inequality is not homogenous I've tried to use $$\lambda _ i = \frac{a_i^{\frac1i -1}}{\sum_{k=1}^n a_k^{\frac1k -1}}$$
this $\lambda$ is from the inequality:
$$ \sum_{i=1}^n \lambda_i a_i \geq \Pi_{i=1}^n a_i^{\lambda_i}$$
the sum of all $\lambda_i$ must be 1. It didn't work and I'm stuck
To be more clear, I need to prove that
$$a_1 + \sqrt{a_2} + \sqrt[3]{a_3} + ... + \sqrt[n]{a_n} \geq \frac{n+1}{2}$$
and we know that $a_1a_2a_3...a_n =1$
|
\begin{align}
&a_1 + \sqrt{a_2} + \sqrt[3]{a_3} + ... + \sqrt[n]{a_n} \\
=& \sum_{k=1}^n \sum_{j=1}^k \frac{1}{k} \sqrt[k]{a_k}\\
\ge & \frac{n(n+1)}{2} \left(\prod_{k=1}^n(\frac{1}{k}\sqrt[k]{a_k})^k\right)^{\frac{2}{n(n+1)}} \\
= & \frac{n(n+1)}{2} \left(\prod_{k=1}^n(\frac{1}{k^k}{a_k})\right)^{\frac{2}{n(n+1)}} \\
= & \frac{n(n+1)}{2} \left(\prod_{k=1}^n\frac{1}{k^k}\right)^{\frac{2}{n(n+1)}} \\
\ge & \frac{n(n+1)}{2} \left(\prod_{k=1}^n\frac{1}{n^k}\right)^{\frac{2}{n(n+1)}} \\
= & \frac{n(n+1)}{2} \left(\frac{1}{n^{\frac{n(n+1)}{2}}}\right)^{\frac{2}{n(n+1)}} \\
= &\frac{n+1}{2}
\end{align}
|
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|
Computing a double gamma-digamma-trigamma series What are your thoughts on this series?
$$\sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{\Gamma (k)^2 \Gamma (n) }{\Gamma (2 k+n)}((\psi ^{(0)}(n)-\psi ^{(0)}(2 k+n)) (\psi ^{(0)}(k)-\psi ^{(0)}(2 k+n))-\psi ^{(1)}(2 k+n)).$$
EDIT: Noting the interest for this series, I wanna add that the series gets reduced to the calculation of $$\int_0^1 \left(\frac{\text{Li}_3(x)}{x^2-2 x+2}+\frac{\text{Li}_3\left(x-x^2\right)}{x^2-2 x+2}-\frac{\text{Li}_3\left(\frac{x}{x-1}\right)}{x^2-2 x+2}-\frac{\text{Li}_3\left(\frac{(x-1) x}{x^2-x+1}\right)}{x^2-2 x+2}-\frac{\text{Li}_2\left(\frac{x}{x-1}\right) \log (1-x)}{x^2-2 x+2}-\frac{\text{Li}_2(x) \log (1-x)}{x^2-2 x+2}+\frac{\text{Li}_2\left(\frac{x}{x-1}\right) \log (x)}{x^2-2 x+2}+\frac{\text{Li}_2\left(\frac{(x-1) x}{x^2-x+1}\right) \log (x)}{x^2-2 x+2}-\frac{\text{Li}_2(x) \log (x)}{x^2-2 x+2}-\frac{\text{Li}_2\left(x-x^2\right) \log (x)}{x^2-2 x+2}-\frac{\text{Li}_2\left(x-x^2\right) \log \left(x^2-x+1\right)}{x^2-2 x+2}-\frac{\text{Li}_2\left(\frac{(x-1) x}{x^2-x+1}\right) \log \left(x^2-x+1\right)}{x^2-2 x+2}-\frac{\log ^3(1-x)}{3 \left(x^2-2 x+2\right)}-\frac{\log ^3\left(x^2-x+1\right)}{3 \left(x^2-2 x+2\right)}-\frac{\log ^2(x) \log (1-x)}{x^2-2 x+2}-\frac{\log ^2(x) \log \left(x^2-x+1\right)}{x^2-2 x+2}+\frac{\pi ^2 \log (1-x)}{6 \left(x^2-2 x+2\right)}+\frac{\pi ^2 \log \left(x^2-x+1\right)}{6 \left(x^2-2 x+2\right)}\right) \, dx$$
where we are pretty familiar with all the stuff in here. One can find the closed form by calculating the integral, a bit long, but it's a nice journey to go.
A 300 points bounty moment: After 2 years and 10 months since the problem has been posed no full solution has been provided. Is it possible to find a slick solution?
|
I'd try to use: $$\displaystyle \int\limits_{0<x+y<1} x^{u-1}y^{u-1}(1-x-y)^{v-1} \,dx\,dy= \frac{\Gamma(u)^2\Gamma(v)}{\Gamma(2u+v)}$$
for reals $u,v \in \mathbb{R}$.
Hence, $$\displaystyle \begin{align}\sum\limits_{n,k=1}^{\infty} \frac{\Gamma(k+u)^2\Gamma(n+v)}{\Gamma(2k+2u+n+v)} &= \int\limits_{0<x+y<1} \frac{(xy)^{u}(1-x-y)^v}{(1-xy)(x+y)}\,dx\,dy\end{align}$$
Since, $\displaystyle \frac{\partial^2}{\partial u\partial v} \left(\frac{\Gamma(u)^2\Gamma(v)}{\Gamma(2u+v)}\right) = \frac{2\Gamma(u)^2\Gamma(v)}{\Gamma(2u+v)}\left((\psi^{(0)}(u) - \psi^{(0)}(2u+v))(\psi^{(0)}(v) - \psi^{(0)}(2u+v)) - \psi^{(1)}(2u+v)\right)$
Hence, $$\begin{align}&2\sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{\Gamma (k)^2 \Gamma (n) }{\Gamma (2 k+n)}((\psi ^{(0)}(n)-\psi ^{(0)}(2 k+n)) (\psi ^{(0)}(k)-\psi ^{(0)}(2 k+n))-\psi ^{(1)}(2 k+n))\\&= \lim\limits_{(u,v) \to (0,0)} \frac{\partial^2}{\partial u\partial v}\int\limits_{0<x+y<1} \frac{(xy)^{u}(1-x-y)^v}{(1-xy)(x+y)}\,dx\,dy\\&= \lim\limits_{(u,v) \to (0,0)} \int\limits_{0<x+y<1} \frac{(xy)^{u}(1-x-y)^v\log (xy)\log(1-x-y)}{(1-xy)(x+y)}\,dx\,dy\end{align}$$
We need to compute the integral and take the limit after that. This is only a partial attempt.
|
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|
Prove that $\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=3/2$
Prove that $$\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=\frac{3}{2}$$
I thought of rewriting $$\cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})$$ as $$\cos^2(90^{\circ}+ (\theta +30^{\circ})) + \cos^2(90^{\circ}-(\theta-30^{\circ}))$$ However I don't seem to get anywhere with this.
Unfortunately I don't know how to solve this question. I would be really grateful for any help or suggestions. Many thanks in advance!
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I am assuming that $\theta$ is in degrees. Then using the following formulas:
\begin{eqnarray}
\cos(x+y)&=&\cos x\cos y-\sin x\sin y\\
\cos(x-y)&=&\cos x\cos y+\sin x\sin y\\
(a-b)^2+(a+b)^2&=&2(a^2+b^2)\\
\cos(120)&=&-\frac12\\
\sin(120)&=&\frac{\sqrt3}{2}\\
\cos^2(x)+\sin^2(x)&=&1
\end{eqnarray}
we have:
\begin{eqnarray}
&& \cos^2(\theta)+\cos^2(\theta+120)+\cos^2(\theta-120) \\[12pt]
&=&\cos^2\theta+[\cos(\theta)\cos(120)-\sin(\theta)\sin(120)]^2\\[8pt]
&&+[\cos(\theta)\cos(120)+\sin(\theta)\sin(120)]^2\\[8pt]
&=&\cos^2(\theta)+2[\cos^2(\theta)\cos^2(120)+\sin^2(\theta)\sin^2(120)]\\[8pt]
&=&\cos^2(\theta)+2\left[\left(-\frac12\right)^2\cos^2(\theta)+\left(\frac{\sqrt3}{2}\right)^2\sin^2(\theta)\right]\\[8pt]
&=&\cos^2(\theta)+2\left[\frac14\cos^2(\theta)+\frac34\sin^2(\theta)\right]\\[8pt]
&=&\cos^2(\theta)+\frac12\cos^2(\theta)+\frac32\sin^2(\theta)\\[8pt]
&=&\left(1+\frac12\right)\cos^2(\theta)+\frac32\sin^2(\theta)\\[8pt]
&=&\frac32\cos^2(\theta)+\frac32\sin^2(\theta)\\[8pt]
&=&\frac32[\cos^2(\theta)+\sin^2(\theta)]\\[8pt]
&=&\frac32\cdot1=\frac32.
\end{eqnarray}
|
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|
Closed form for multiplicative recurrence relation In this StackOverflow question, I found an interesting recurrence relation:
$$f(n) = \begin{cases} 1 & n \leq 2 \\ nf(n-1) + (n-1)f(n-2) & \text{otherwise.}\end{cases}$$
I plugged it into Wolfram Alpha, and it gives me the solution:
$$f(n) = \frac{2~\Gamma(n+3) - 5~!(n+2)}{n+1}$$
How would a human being go about solving something like this? Could I use generating functions in some clever way?
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Note that $\Gamma(n+3)=(n+2)!$.
$$f(n) = \frac{2\Gamma(n+3) - 5!(n+2)}{n+1}=\frac{2(n+2)! - 5 \cdot !(n+2)}{n+1}$$
Once one has this form, it is not hard using induction:
\begin{align*} f(n+1)&=(n+1)f(n)+nf(n-1) \\ &=\frac{2(n+2)! - 5 \cdot !(n+2)}{n+1}(n-2)+\frac{2(n+1)! - 5 \cdot !(n+1)}{n}n \\ &= 2(n+2)! - 5 \cdot !(n+2) + 2(n+1)! - 5 \cdot !(n+1) \\ &= 2(n+2)(n+1)! - 5 \cdot !(n+2) + 2(n+1)! - 5 \cdot !(n+1) \\ &= 2(n+3)(n+1)! - 5 \cdot !(n+2) - 5 \cdot !(n+1) \\ &= 2(n+3)(n+1)! - 5 \cdot (!(n+2)+ !(n+1)) \\&= \frac{[2(n+3)(n+1)! - 5 \cdot (!(n+2)+ !(n+1))](n-2)}{n-2} \\&= \frac{2(n+3)(n-2)(n+1)! - 5(n-2) \cdot (!(n+2)+ !(n+1))}{n-2} \\&= \frac{2(n+3)! - 5 \cdot !(n+3)}{n+2}\end{align*}
The last step uses the recurrence relation given here.
To find such formula, just notice that the closed form is the sum of a lot of factorials, and you could probably find it by writing it as factorials while doing some small cases.
|
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Definite Integration with Trigonometric Substitution I'm working on a question that involves using trigonometric substitution on a definite integral that will later use u substitution but I am not sure how to go ahead with this.
$$\int_1^2\frac1{x^2\sqrt{4x^2+9}}dx$$
My first step was to use $\sqrt{a^2+x^2}$ as $x=a\tan\theta$ to get...
$$2x=3\tan\theta :x=\frac32\tan\theta$$
$$dx=\frac32\sec^2\theta$$
Substituting:
$$\int\frac{\frac32\sec^2\theta}{\frac94\tan^2\theta\sqrt{9\tan^2\theta+9}}$$
The problem here is how do I change the limit it goes to?
$$\frac43=\tan\theta$$
and
$$\frac23=\tan\theta$$
Following DR.MV's answer so far..
$$\frac29\int_{\tan^-1(\frac23)}^{\tan^-1(\frac43)}\frac{\sec^2\theta}{\tan^2\theta\sqrt{9\sec^2\theta}}d\theta$$
$$=\frac29\int_{\tan^-1(\frac23)}^{\tan^-1(\frac43)}\frac{\sec\theta}{\tan^2\theta}d\theta$$
$$=\frac29\int_{\tan^-1(\frac23)}^{\tan^-1(\frac43)}\frac{\cos\theta}{\sin^2\theta}d\theta$$
Now $u=\sin\theta$ so $du=\cos\theta d\theta$
$$=\frac29\int_{?}^{?}\frac{1}{u^2}du$$
This is where I am stuck now...
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\begin{align*}\int\frac{\frac32\sec^2\theta\,\mathrm d\mkern1.5mu\theta}{\frac94\tan^2\theta\sqrt{9\tan^2\theta+9}}&=\frac29\int\frac{\mathrm d\mkern1.5mu\theta}{\sin^2\theta\sqrt{1+\tan^2\theta}}=\frac29\int\frac{\lvert\cos\theta\rvert\,\mathrm d\mkern1.5mu\theta}{\sin^2\theta}\\[1ex]
&=\frac29\int\frac{\cos\theta\,\mathrm d\mkern1.5mu\theta}{\sin^2\theta}\qquad\text{since}\enspace 0\le\theta<\dfrac\pi2\\[1ex]
&=-\frac2{9\sin\theta}
\end{align*}
Some trigonometry will let you determine the bounds for $\sin\theta\;$ from the bounds for $\tan\theta$: since $0\le\theta<\dfrac\pi2$, we have:
$\cos\theta=\dfrac1{\sqrt{1+\tan^2\theta}}$, hence $$\sin\theta=\tan\theta\cos\theta=\frac{\tan\theta}{\sqrt{1+\tan^2\theta}}=\frac{\cfrac{2x}3}{\sqrt{1+\cfrac{4x^2}9}}=\frac{2x}{\sqrt{4x^2+9}}$$
so that the indefinite integral is:
$$-\frac{\sqrt{4x^2+9}}{9x}.$$
|
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Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$ If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the reverse inequality). Need some stronger inequality. Thanks.
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Let $a=\frac{x^2}{yz}$, $b=\frac{y^2}{xz}$ and $c=\frac{z^2}{xy}$, where $x$, $y$ and $z$ are positives.
Hence, by Holder and AM-GM we obtain:
$$\sum_{cyc}\sqrt{\frac{a}{a+8}}=\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{x}{\sqrt{x^2+8yz}}\right)^2\sum\limits_{cyc}x(x^2+8yz)}{\sum\limits_{cyc}x(x^2+8yz)}}\geq$$
$$\geq\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}(x^3+8xyz)}}\geq\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}(x^3+3x^2y+3x^2z+2xyz)}}=1.$$
Done!
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Formulae for sequences Given that for $1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4}$
deduce that $(n+1)^3 + (n+2)^3 +\cdots+ (2n)^3 = \frac{n^2(3n+1)(5n+3)}{4}$
So far:
the sequence $(n+1)^3 + (n+2)^3 +\cdots+ (2n)^3$ gives $2^3 + 3^3 + 4^3 +\cdots,$ when n=1.
The brackets in the formula for the second sequence are $2n$ and $4n+2$ bigger than $(n+1)$ in the original i.e. $(3n+1)(5n+3).$
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Adding the two sums, you will find the sum for $2n$.
Indeed,
$$n^2(n+1)^2+n^2(3n+1)(5n+3)=n^2(16n^2+16n+4)=(2n)^2(2n+1)^2.$$
|
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if $(1-a)(1-b)(1-c)(1-d) = \frac{9}{16}$ then minimum integer value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = ?$ Given $a,b,c,d > 0$, how do we find the minimum integer value of $n=\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}$ such that $(1-a)(1-b)(1-c)(1-d) = \frac{9}{16}$.
|
If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\leq 1$, then $a,b,c,d>1$ and
$$\begin{align}
0\leq abcd-bcd-cda-dab-abc=&(1-a)(1-b)(1-c)(1-d)-1+(a+b+c+d)
\\&-(ab+ac+ad+bc+bd+cd)
\\
=&\frac{9}{16}-1-a(b-1)-b(c-1)-c(d-1)-d(a-1)
\\
&-ac-bd
\\
<&-\frac{7}{16}\,,
\end{align}
$$
which is a contradiction. Hence, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}>1$. Now, $$(a,b,c,d)=\left(\frac{161-\sqrt{8257}}{64},2,3,\frac{161+\sqrt{8257}}{64}\right)$$
is a solution to $(a-1)(b-1)(c-1)(d-1)=\frac{9}{16}$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=2$. Ergo, the minimum possible integer value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$ is $2$.
I think it is a nice question to ask whether there exists a solution $(a,b,c,d)$ with $a,b,c,d>0$ to $(a-1)(b-1)(c-1)(d-1)=\frac{9}{16}$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=t$ for any $t>1$. I conjecture that the answer is positive.
|
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Polynomial of $11^{th}$ degree Let $f(x)$ be a polynomial of degree $11$ such that $f(x)=\frac{1}{x+1}$,for $x=0,1,2,3.......,11$.Then what is the value of $f(12)?$
My attempt at this is:
Let $f(x)=a_0+a_1x+a_2x^2+a_3x^3+......+a_{11}x^{11}$
$f(0)=\frac{1}{0+1}=1=a_0$
$f(1)=\frac{1}{1+1}=\frac{1}{2}=a_0+a_1+a_2+a_3+......+a_{11} $
$f(2)=\frac{1}{2+1}=\frac{1}{3}=a_0+2a_1+4a_2+8a_3+......+2^{11}a_{11} $
.
.
.
$f(11)=\frac{1}{11+1}=\frac{1}{12}=a_0+11a_1+11^2a_2+11^3a_3+......+11^{11}a_{11} $
for calculating $f(12)$, I need to calculate $a_0,a_1,a_2,....,a_11$ but I could solve further.Is my approach right,how can I solve further or there is another right way to solve it.
$(A)\frac{1}{13}$
$(B)\frac{1}{12}$
$(C)0 $
$(D)\frac{1}{7}$
which one is correct answer?
|
HINT:
Let $(x+1)f(x)=1+A\prod_{r=0}^{11}(x-r)$ where $A$ is an arbitrary constant
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|
Solve $10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$ How to solve the following equation?
$$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$$
My attempt:
$$ 10x^4 - (7x^2+1)(x^2+x+1)=0$$
Thats all i can
Update
Tried to open brakets and simplify:
$$(7x^2+1)(x^2+x+1) = 7x^4+7x^3+7x^2+x^2+x+1=7x^4+7x^3+8x^2+1 $$
$$10x^4 - (7x^2+1)(x^2+x+1)= 3x^4-7x^3-8x^2-1=0 $$
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Divide by $(x^2 + x + 1)^2$, the equation becomes: $$10\frac{x^4}{(x^2 + x + 1)^2} - 7\frac{x^2}{x^2 + x + 1} + 1 = 0$$ Let $z = \frac{x^2}{x^2 + x + 1}$. The equation now is $$10z^2 - 7z + 1 = 0$$ Solving it like an ordinary quadratic equation on $z$ you get at most two roots $z_{1,2}$. Then let $\frac{x^2}{x^2 + x + 1} = z_1$, or, equivalently, $$x^2 = z_1 (x^2+x+1)$$ It is a quadratic equation in $x$. Similary for $z_2$. Solutions to this two equations are solutions to the initial problem.
|
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|
Am I getting the right answer for the integral $I_n= \int_0^1 \frac{x^n}{\sqrt {x^3+1}}\, dx$?
Let $I_n= \int_0^1 \dfrac{x^n}{\sqrt {x^3+1}}\, dx$. Show that $(2n-1)I_n+2(n-2)I_{n-3}=2 \sqrt 2$ for all $n \ge 3$. Then compute $I_8$.
I get an answer for $I_8={{2 \sqrt 2} \over 135}(25-16 \sqrt 2)$, could somebody please check against my answer and see if I made a mistake.
|
\begin{align*}I_n&=\int_0^1 \dfrac{x^n}{\sqrt{x^3+1}}dx=\int_0^1 \dfrac{x^{n-3}(x^3+1-1)}{\sqrt{x^3+1}}dx = \int_0^1 x^{n-3}\sqrt{x^3+1}dx - \int_0^1 \dfrac{x^{n-3}}{\sqrt{x^3+1}}dx\\ &= \int_0^1 x^{n-3}\sqrt{x^3+1}dx-I_{n-3}\end{align*}This integral is handled with integration by parts:
$$\int_0^1 x^{n-3}\sqrt{x^3+1}dx=\frac{x^{n-2}}{n-2}\sqrt{x^3+1}|_0^1-\frac{3}{2(n-2)}\int_0^1 \dfrac{x^{n}}{\sqrt{x^3+1}}dx=\dfrac{\sqrt{2}}{n-2}-\frac{3}{2(n-2)}I_n$$Therefore $$I_n=\dfrac{\sqrt{2}}{n-2}-\frac{3}{2(n-2)}I_n-I_{n-3}$$ or $$(n-2+\frac{3}{2})I_n+(n-2)I_{n-3}=\sqrt{2}\tag{1}$$Which is equivalent to your formula, so I would guess you're correct, though I haven't computed $I_2$ to check if $I_8$ is correct.
|
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|
Prove that $\tan\alpha =\tan^{2}\frac{A}{2}.\tan\frac{B-C}{2}$ Given a triangle ABC with the sides $AB < AC$ and $AM, AD$ respectively median and bisector of angle $A$. Let $\angle MAD = \alpha$. Prove that $$\tan\alpha =\tan^{2}\frac{A}{2}\cdot \tan\frac{B-C}{2}$$
|
Firstly, in figure 1,
Using Napier's Analogy in $\Delta ABC$, we have,
$$\tan \left(\frac{B-C}{2}\right) = \frac{b-c}{b+c} \cdot \cot \left( \frac{A}{2}\right)$$
Now, in figure 2,
Using Sine Law in $\Delta ABM$, we have,
$$\dfrac{x}{\sin \left(\frac{A}{2} + \alpha \right)}=\dfrac{c}{ \sin(\angle BMA)}$$
Similarly, in $\Delta ACM$,
$$\dfrac{x}{\sin \left(\frac{A}{2} - \alpha \right)}=\dfrac{b}{ \sin(\angle CMA)}$$
Also, $\sin(\angle CMA)= \sin(\pi-\angle BMA)= \sin(\angle BMA) = \omega $ (let)
$\therefore b = \dfrac{\omega x}{\sin\left(\frac{A}{2}-\alpha\right)}$
$\text{and}$
$c=\dfrac{\omega x}{\sin\left(\frac{A}{2}+\alpha\right)}$
Putting the values of $b$ and $c$ in the equation of Napier's Analogy, we get (after trivial cancellations),
$$\tan \left(\dfrac{B-C}{2}\right) = \dfrac{\sin\left(\frac{A}{2}+\alpha\right)-\sin\left(\frac{A}{2}-\alpha\right)}{\sin\left(\frac{A}{2}+\alpha\right)+\sin\left(\frac{A}{2}-\alpha\right)} \cdot \cot \left( \dfrac{A}{2}\right)$$
$$\implies \tan \left(\dfrac{B-C}{2}\right) = \dfrac{2 \cos\left(\frac{A}{2}\right) \cdot \sin \alpha}{2\sin\left(\frac{A}{2}\right) \cdot \cos \alpha} \cdot \cot \left( \dfrac{A}{2}\right)$$
$$\implies \tan \alpha = \tan^2 \left( \dfrac{A}{2}\right) \cdot \tan \left(\dfrac{B-C}{2}\right) $$
Q.E.D.
|
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|
How to solve this equality? [3] $$4x^2 - 6x^4 + \frac{8x^6 - 2x^2 - \frac{1}{x^2}}{16} = 0$$
The equation has a strange look, and as such is probably as it should not be solved. Maybe the roots of trigonometric functions are expressed in terms of angles species $\frac{\pi}{n}$?
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$$4x^2-6x^4+\frac{8x^6-2x^2-\frac{1}{x^2}}{16}=0$$
$$4x^2-6x^4+\frac{\frac{8x^8-2x^4-1}{x^2}}{16}=0$$
$$4x^2-6x^4+\frac{8x^8-2x^4-1}{16x^2}=0$$
$$\frac{64x^4-96x^6+8x^8-2x^4-1}{16x^2}=0$$
$$8x^8-96x^6+64x^4-2x^4-1=0$$
|
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|
Struggling with an inequality: $ \frac{1}{\sqrt[n]{1+m}} + \frac{1}{\sqrt[m]{1+n}} \ge 1 $ Prove that for every natural numbers, $m$ and $n$, this inequality holds:
$$
\frac{1}{\sqrt[n]{1+m}} + \frac{1}{\sqrt[m]{1+n}} \ge 1
$$
I tried to use Bernoulli's inequality, but I can't figure it out.
|
The AM-GM inequality says that for $x,y\ge0$ and $0\le a\le1$,
$$
ax+(1-a)y\ge x^ay^{1-a}\tag{1}
$$
substituting $x\mapsto x/a$ and $y\mapsto y/(1-a)$ yields
$$
x+y\ge\frac{x^ay^{1-a}}{a^a(1-a)^{1-a}}\tag{2}
$$
Therefore, with $x=(m+1)^{-1/n}$, $y=(n+1)^{-1/m}$, and $a=\frac n{m+n}$, we get
$$
\begin{align}
(m+1)^{-1/n}+(n+1)^{-1/m}
&\ge\frac{[(m+1)(n+1)]^{-\frac1{m+n}}}{\left(\frac m{m+n}\right)^{\frac m{m+n}}\left(\frac n{m+n}\right)^{\frac n{m+n}}}\\[6pt]
&=\frac{m+n}{\left[(m+1)m^m(n+1)n^n\right]^{\frac1{m+n}}}\\[12pt]
&\ge1\tag{3}
\end{align}
$$
The last step in $(3)$ is equivalent to
$$
(m+n)^{m+n}\ge(m+1)(n+1)m^mn^n\tag{4}
$$
Note that $(4)$ is true when $m=n=1$. Suppose that $(4)$ is true for some $m,n$, then as $m$ or $n$ is increased by $1$, the left side of $(4)$ increases by a factor of
$$
\frac{(m+n+1)^{m+n+1}}{(m+n)^{m+n}}
=\color{#C00000}{(m+n+1)}\color{#00A000}{\left(1+\frac1{m+n}\right)^{m+n}}\tag{5}
$$
and if $n$ is increased by $1$ (and similarly for $m$), the right side of $(4)$ increases by a factor of
$$
\frac{(n+2)(n+1)^{n+1}}{(n+1)n^n}
=\color{#C00000}{(n+2)}\color{#00A000}{\left(1+\frac1n\right)^n}\tag{6}
$$
Since $m,n\ge1$, we have that $\color{#C00000}{m+n+1\ge n+2}$, and Bernoulli's Inequality ensures that
$\color{#00A000}{\left(1+\frac1{m+n}\right)^{m+n}\ge\left(1+\frac1n\right)^n}$.
Comparing $(5)$ and $(6)$ ensures that inequality $(4)$ remains valid for all $m,n\ge1$.
|
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|
Prove using mathematical induction that $x^{2n} - y^{2n}$ is divisible by $x+y$ Prove using mathematical induction that
$(x^{2n} - y^{2n})$ is divisible by $(x+y)$.
Step 1: Proving that the equation is true for $n=1 $
$(x^{2\cdot 1} - y^{2\cdot 1})$ is divisible by $(x+y)$
Step 2: Taking $n=k$
$(x^{2k} - y^{2k})$ is divisible by $(x+y)$
Step 3: proving that the above equation is also true for $(k+1)$
$(x^{2k+2} - y^{2k+2})$ is divisible by $(x+y)$.
Can anyone assist me what would be the next step? Thank You in advance!
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Step 1: putting $n=1$, we get $$x^{2n}-y^{2n}=x^2-y^2=(x-y)(x+y)$$ above number is divisible by $(x+y)$. Hence the statement is true $n=1$
step 2: assuming that for $n=m$, $(x^{2n}-y^{2n})$ is divisible by $(x+y)$ then we have $$(x^{2m}-y^{2m})=k(x+y) \tag 1$$
Where, $k$ is an integer
step 3: putting $n=m+1$ we get $$x^{2(m+1)}-y^{2(m+1)}$$ $$=x^{2m+2}-y^{2m+2}$$
$$=x^{2m+2}-x^{2m}y^2-y^{2m+2}+x^{2m}y^2$$ $$=(x^{2m+2}-x^{2m}y^2)+(x^{2m}y^2-y^{2m+2})$$ $$=x^{2m}(x^2-y^2)+y^2(x^{2m}-y^{2m})$$ Setting the value of $(x^{2m}-y^{2m})$ from (1), we get $$x^{2m}(x-y)(x+y)+y^2k(x+y)$$ $$=k(x^{2m}(x-y)+y^2)(x+y)$$ It is clear that the above number is divisible by $(x+y)$. Hence the statement holds for $n=m+1$
Thus $(x^{2n}-y^{2n})$ is divisible by $(x+y)$ for all positive integers $\color{blue}{n\geq 1}$
|
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|
How to integrate ${x^3}/(x^2+1)^{3/2}$? How to integrate $$\frac{x^3}{(x^2+1)^{3/2}}\ \text{?}$$
I tried substituting $x^2+1$ as t, but it's not working
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Alternative approach:
Let $x=\tan\theta$, $dx=\sec^2\theta d\theta$
\begin{align}
\int\frac{x^3dx}{(x^2+1)^{3/2}}&=\int\frac{\tan^3\theta\cdot\sec^2\theta d\theta}{\sec^3\theta}\\&=\int\frac{\sin^3\theta d\theta}{\cos^2\theta}\\&=-\int\frac{\sin^2\theta d\cos\theta}{\cos^2\theta}\\&=\int1-\sec^2\theta d\cos\theta\\&=\cos\theta+\sec\theta+C\\&=\frac1{\sqrt{1+x^2}}+\sqrt{1+x^2}+C
\end{align}
|
{
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|
Is $\left(45+29\sqrt{2}\right)^{1/3} + \left(45-29\sqrt{2}\right)^{1/3}$ an integer? The problem is the following:
Prove that this number
$$x = \left(45+29\sqrt{2}\right)^{1/3} + \left(45-29\sqrt{2}\right)^{1/3}$$
is an integer. Show which integer it is.
I thought that it has some relations with something like complex numbers such as the set $N(\sqrt{2})$ or the some kind of integer polynomial which has root $x$ and therefore show that it has integer solutions.
|
Let $u=\sqrt[3]{45+29\sqrt2}$ and $v=\sqrt[3]{45-29\sqrt2}$; so, $x=u+v$. We have
$$
u^3 + v^3=90,\\
uv = \sqrt[3]{45+29\sqrt2}\cdot \sqrt[3]{45-29\sqrt2} = \sqrt[3]{45^2-29^2\cdot2} = \sqrt[3]{343} = 7
$$
But
$$
u^3 + v^3 = (u+v)(u^2 - uv + v^2) = (u+v)((u+v)^2 - 3uv),
$$
and
$$
90 = x(x^2 - 21).
$$
Since $x$ must be integer, check divisors of $90$. $x^2 - 21 > 0$, so $x \ge 5$. Try $x=6$:
$$
90 = 6(6^2-21) = 6\cdot 15.
$$
|
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|
Find all values that solve the equation For which values a, the equation
$$ a\sin{x}+(a+1)\sin^2{\frac{x}{2}} + (a-1)\cos^2{\frac{x}{2}} =1 $$
has a solution?
My idea: I think it's possible to factorize equation or reduce equation to the form like: $a(\sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}}) =1 $
Let's go:
$$ 2a\sin{\frac{x}{2}}\cos{\frac{x}{2}} + asin^2{\frac{x}{2}} + sin^2{\frac{x}{2}} + a\cos^2{\frac{x}{2}} - \cos^2{\frac{x}{2}} = 1$$
$$ a\left(sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right)^2 = 1 - \sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}}$$
$$ a\left(\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right)^2 =2\cos^2{\frac{x}{2}}$$
I can't finish...
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$$ a\sin x + a(\sin^2 (x/2) + \cos^2 (x/2)) + \sin^2 (x/2) - \cos^2 (x/2) =1$$
$$ a\sin x + a(\sin^2 (x/2) + \cos^2 (x/2)) + 1 - 2\cos^2 (x/2) =1$$
$$ a \sin x + a - \cos x =1$$
|
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|
Find the equation of the circle. Find the equation of the circle whose radius is $5$ which touches the circle $x^2 + y^2 - 2x -4y - 20 = 0$ externally at the point $(5,5)$
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HINT:
$(x-a)^2+(y-b)^2=5^2$ will touch $(x-1)^2+(y-2)^2=5^2$
iff $5+5=\sqrt{(a-1)^2+(b-2)^2}$
Again, $(a,b), (1,2), (5,5)$ are collinear.
So, we have two equations with two unknowns
|
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|
The value of the definite integral The value of the definite integral $\displaystyle\int\limits_0^\infty \frac{\ln x}{x^2+4} \, dx$ is
(A) $\dfrac{\pi \ln3}{2}$ (B) $\dfrac{\pi \ln2}{3}$ (C) $\dfrac{\pi \ln2}{4}$ (D) $\dfrac{\pi \ln4}{3}$
I tried using integration by parts,
\begin{align}
& \int_0^\infty \frac{\ln x}{x^2+4}dx = \ln x\int_0^\infty \frac{1}{x^2+4}dx-\int_0^\infty \left(\frac{d}{dx}\ln x\int_0^\infty \frac{1}{x^2+4}\right) \, dx \\[10pt]
= {} & \left[\ln x \frac{1}{2}\tan^{-1}\frac{x}{2}\right]-\int_0^\infty \frac{1}{x}\frac{1}{2}\tan^{-1}\frac{x}{2} \, dx
\end{align}
I could not move ahead. Can someone help me to get final answer?
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$\bf{My\; Solution::}$ Let $\displaystyle I = \int_{0}^{\infty}\frac{\ln(x)}{x^2+4}dx\;,$ Now Let $x=2t\;,$ Then $dx = 2dt$
and Changing Limits, we get
$$\displaystyle I = \int_{0}^{\infty}\frac{\ln(2t)}{4t^2+4}\times 2dt = \frac{1}{2}\int_{0}^{\infty}\frac{\ln(2t)}{t^2+1}dt=\frac{1}{2}\int_{0}^{\infty}\frac{\ln(2)}{t^2+1}dt+\frac{1}{2}\int_{0}^{\infty}\frac{\ln(t)}{t^2+1}dt$$
So $$\displaystyle I = \frac{\ln(2)}{2}\int_{0}^{\infty}\frac{1}{1+t^2}dt+\frac{1}{2}J = \frac{\ln(2)}{2}\cdot \frac{\pi}{2}+\frac{1}{2}J...........(1)$$
Now For Calculation of $$\displaystyle J = \frac{1}{2}\int_{0}^{\infty}\frac{\ln(t)}{t^2+1}dt\;,$$ Put $\displaystyle t=\frac{1}{u}$ and $\displaystyle dt = -\frac{1}{u^2}du$
and Changing Limits, We get $$\displaystyle J=-\frac{1}{2}\int_{0}^{\infty}\frac{\ln(u)}{1+u^2}du = -\frac{1}{2}\int_{0}^{\infty}\frac{\ln(t)}{1+t^2}dt = -J$$
So We get $J = -J\Rightarrow J=0\;,$ Now Put into $(1)$ equation, We get $$\displaystyle I=\int_{0}^{\infty}\frac{\ln(x)}{x^2+4}dx = \frac{\pi\cdot \ln(2)}{4}.$$
|
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|
Some help on trigonometric equation So I have $\sin^3x = \frac 34 \sin x$.
Can you expand so the answer is either $\sin x(\sin^2x +\frac 34)$ which leads to
the answer $\frac 12 + 2n\pi$ or that $\sin^3x = \frac 14(3\sin x-\sin^3x) - \frac 34\sin x$ which leads to the answer $0 + 2n \pi$.
Is that correct by any chance?
|
Notice, we have $$\sin^3x = \frac 34 \sin x$$ $$\sin^3x -\frac 34 \sin x=0$$ $$\sin x\left(\sin^2 x-\frac{3}{4}\right)=0$$ $$\text{if}\ \sin x=0\ \implies \color{blue}{x=n\pi}$$ $$\text{if}\ \sin^2 x-\frac{3}{4}\iff \sin^2x=\left(\frac{\sqrt{3}}{2}\right)^2 \iff \sin^2x=\left(\sin \frac{\pi}{3}\right)^2\ \implies \color{blue}{x=n\pi\pm \frac{\pi}{3}}$$ Where, $n$ is an integer.
|
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|
Matrix $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ to a large power Compute
$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{99}$
What is the easier way to do this other than multiplying the entire thing out?
Thanks
|
Consider the case of a general $2 \times 2$ matrix times $\left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right)$:
$\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \times \left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right) = \left( \begin{matrix} a & a + b \\ c & c + d \end{matrix} \right) $.
Thus, we see that multiplying any $2 \times 2$ matrix by $\left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right)$ once leaves the entries in the first column unchanged and increments the entries in the second column by the corresponding entries from the first column. Since our lower-left entry was zero and the upper-left entry was one, this has the effect of simply incrementing the upper-right entry by one but leaving the rest unchanged.
Therefore, since we begin with $\left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right)$ and multiply it by $\left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right)$ 98 times, we see that $\left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right)^{99} = \left( \begin{matrix} 1 & 99 \\ 0 & 1 \end{matrix} \right) $.
|
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|
Show by induction that $F_n \geq 2^{0.5 \cdot n}$, for $n \geq 6$ I have the following problem:
Show by induction that $F_n \geq 2^{0.5 \cdot n}$, for $n \geq 6$
Where $F_n$ is the $nth$ Fibonacci number.
Proof
Basis
$n = 6$.
$F_6 = 8 \geq 2^{0.5 \cdot 6} = 2^{\frac{6}{2}} = 2^3 = 8$
Induction hypothesis
Assume $F_n \geq 2^{\frac{n}{2}}$, for some $n \geq 6$.
Inductive step
Lets shows that $F_{n+1} \geq 2^{\frac{n + 1}{2}}$.
We know that
*
*$F_{n + 1} = F_n + F_{n - 1}$
*$2^{\frac{n + 1}{2}} > 2^{\frac{n}{2}}$
*$2^{\frac{n + 1}{2}} = 2^{\frac{n}{2} + \frac{1}{2}} = 2^{\frac{n}{2}} \cdot 2^{\frac{1}{2}} \leq 2^{\frac{n}{2}} \cdot 2 = 2^{\frac{n}{2}} + 2^{\frac{n}{2}}$
Since we have assumed that $F_n \geq 2^{\frac{n}{2}}$, then $$F_n + F_{n - 1} = F_{n + 1} \geq 2^{\frac{n}{2}} + F_{n - 1} \geq 2^{\frac{n}{2}} + F_{n - 1} + 2^{\frac{n}{2}} - F_n = 2^{\frac{n}{2}} + 2^{\frac{n}{2}} + F_{n - 1} - F_n$$
The last inequality is true because $2^{\frac{n}{2}} - F_n$ is negative or $0$, since $F_n \geq 2^{\frac{n}{2}}$.
I have tried a lot of things, but I cannot figure out how to proceed and how to conclude that indeed $F_{n + 1} \geq 2^{\frac{n}{2}} \cdot 2^{\frac{1}{2}}$. I feel really stupid after trying for a long time to do this alone and not managing to do it.
|
We have that $F_n>F_{n-1}$ then $$F_{n+1}=F_n+F_{n-1}>2F_{n-1}>2\cdot2^{(n-1)/2}=2^{(n+1)/2}$$
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.